NCERT Solutions · Class 9 Mathematics Exploring Algebraic Identities

Read the shape, then choose the identity. A square of a binomial calls for \(\displaystyle (a\pm b)^{2}\); a product of a sum and a difference of the same two terms calls for \(\displaystyle (a+b)(a-b)=a^{2}-b^{2}\); a square of three terms calls for \(\displaystyle (a+b+c)^{2}\); a cube calls for \(\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}\); and \(\displaystyle (a-b)(a^{2}+ab+b^{2})=a^{3}-b^{3}\).(i) \(\displaystyle (-3x+4)^{2}\). Take \(\displaystyle a=-3x\), \(\displaystyle b=4\): \(\displaystyle a^{2}=9x^{2}\), \(\displaystyle 2ab=2(-3x)(4)=-24x\), \(\displaystyle b^{2}=16\). \[(-3x+4)^{2}=9x^{2}-24x+16 \](ii) \(\displaystyle (2s+7)(2s-7)\) — a sum times the matching difference, so \(\displaystyle a^{2}-b^{2}\) with \(\displaystyle a=2s\), \(\displaystyle b=7\). \[(2s+7)(2s-7)=4s^{2}-49 \](iii) \(\displaystyle \left(p^{2}+\frac{1}{2}\right)\left(p^{2}-\frac{1}{2}\right)\), with \(\displaystyle a=p^{2}\), \(\displaystyle b=\frac{1}{2}\). \[=p^{4}-\tfrac{1}{4} \](iv) \(\displaystyle (2n+7)(2n-7)\) — the same as (ii) with \(\displaystyle n\) in place of \(\displaystyle s\). \[=4n^{2}-49 \](v) \(\displaystyle (s-2t)(s^{2}+2st+4t^{2})\). This is \(\displaystyle (a-b)(a^{2}+ab+b^{2})\) with \(\displaystyle a=s\), \(\displaystyle b=2t\): note \(\displaystyle ab=2st\) and \(\displaystyle b^{2}=4t^{2}\), exactly the bracket given. \[(s-2t)(s^{2}+2st+4t^{2})=s^{3}-8t^{3} \](vi) \(\displaystyle \left(\frac{1}{2r}-4r\right)^{2}\), with \(\displaystyle a=\frac{1}{2r}\), \(\displaystyle b=4r\): \(\displaystyle a^{2}=\frac{1}{4r^{2}}\), \(\displaystyle 2ab=2\cdot\frac{1}{2r}\cdot 4r=4\), \(\displaystyle b^{2}=16r^{2}\). \[\left(\tfrac{1}{2r}-4r\right)^{2}=\frac{1}{4r^{2}}-4+16r^{2} \](vii) \(\displaystyle (-3m+4k-l)^{2}\). Use \(\displaystyle (a+b+c)^{2}\) with \(\displaystyle a=-3m\), \(\displaystyle b=4k\), \(\displaystyle c=-l\): \[a^{2}+b^{2}+c^{2}=9m^{2}+16k^{2}+l^{2}, \] \[2ab=2(-3m)(4k)=-24km,\quad 2bc=2(4k)(-l)=-8kl,\quad 2ca=2(-l)(-3m)=6lm. \] \[(-3m+4k-l)^{2}=9m^{2}+16k^{2}+l^{2}-24km-8kl+6lm \](viii) \(\displaystyle \left(x-\frac{1}{3}y\right)^{3}\). Use \(\displaystyle (a-b)^{3}\) with \(\displaystyle a=x\), \(\displaystyle b=\frac{y}{3}\): \[a^{3}=x^{3},\quad 3a^{2}b=3x^{2}\cdot\tfrac{y}{3}=x^{2}y,\quad 3ab^{2}=3x\cdot\tfrac{y^{2}}{9}=\tfrac{xy^{2}}{3},\quad b^{3}=\tfrac{y^{3}}{27}. \] \[\left(x-\tfrac{1}{3}y\right)^{3}=x^{3}-x^{2}y+\frac{xy^{2}}{3}-\frac{y^{3}}{27} \](ix) \(\displaystyle \left(\frac{7}{2}k-\frac{2}{3}m\right)^{3}\), with \(\displaystyle a=\frac{7k}{2}\), \(\displaystyle b=\frac{2m}{3}\): \[a^{3}=\frac{343k^{3}}{8},\qquad 3a^{2}b=3\cdot\frac{49k^{2}}{4}\cdot\frac{2m}{3}=\frac{49k^{2}m}{2}, \] \[3ab^{2}=3\cdot\frac{7k}{2}\cdot\frac{4m^{2}}{9}=\frac{14km^{2}}{3},\qquad b^{3}=\frac{8m^{3}}{27}. \] \[\left(\tfrac{7}{2}k-\tfrac{2}{3}m\right)^{3}=\frac{343}{8}k^{3}-\frac{49}{2}k^{2}m+\frac{14}{3}km^{2}-\frac{8}{27}m^{3} \]Answers: (i) \(\displaystyle 9x^{2}-24x+16\); (ii) \(\displaystyle 4s^{2}-49\); (iii) \(\displaystyle p^{4}-\frac{1}{4}\); (iv) \(\displaystyle 4n^{2}-49\); (v) \(\displaystyle s^{3}-8t^{3}\); (vi) \(\displaystyle 16r^{2}-4+\frac{1}{4r^{2}}\); (vii) \(\displaystyle 9m^{2}+16k^{2}+l^{2}-24km-8kl+6lm\); (viii) \(\displaystyle x^{3}-x^{2}y+\frac{xy^{2}}{3}-\frac{y^{3}}{27}\); (ix) \(\displaystyle \frac{343}{8}k^{3}-\frac{49}{2}k^{2}m+\frac{14}{3}km^{2}-\frac{8}{27}m^{3}\).

Turn each number into a round number plus or minus a small one. For products, check whether the two numbers are equally spaced about a round number — then \(\displaystyle (a+b)(a-b)=a^{2}-b^{2}\) applies. For cubes, use \(\displaystyle (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\) or \(\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}\).(i) \(\displaystyle 17\times 21\): both are \(\displaystyle 2\) away from \(\displaystyle 19\). \[(19-2)(19+2)=19^{2}-2^{2}=361-4=357 \](ii) \(\displaystyle 104\times 96\): both are \(\displaystyle 4\) away from \(\displaystyle 100\). \[(100+4)(100-4)=10000-16=9984 \](iii) \(\displaystyle 24\times 16\): both are \(\displaystyle 4\) away from \(\displaystyle 20\). \[(20+4)(20-4)=400-16=384 \](iv) \(\displaystyle 147^{3}=(150-3)^{3}\), with \(\displaystyle a=150\), \(\displaystyle b=3\): \[a^{3}=3375000,\quad 3a^{2}b=3(22500)(3)=202500,\quad 3ab^{2}=3(150)(9)=4050,\quad b^{3}=27. \] \[147^{3}=3375000-202500+4050-27=3176523 \](v) \(\displaystyle 199^{3}=(200-1)^{3}\): \[=8000000-3(40000)(1)+3(200)(1)-1=8000000-120000+600-1=7880599 \](vi) \(\displaystyle 127^{3}=(130-3)^{3}\), with \(\displaystyle a=130\), \(\displaystyle b=3\): \[a^{3}=2197000,\quad 3a^{2}b=3(16900)(3)=152100,\quad 3ab^{2}=3(130)(9)=3510,\quad b^{3}=27. \] \[127^{3}=2197000-152100+3510-27=2048383 \](vii) \(\displaystyle (-107)^{3}\). An odd power of a negative number is negative, so \(\displaystyle (-107)^{3}=-\left(107^{3}\right)\). Now \(\displaystyle 107^{3}=(100+7)^{3}\): \[=1000000+3(10000)(7)+3(100)(49)+343=1000000+210000+14700+343=1225043. \] \[(-107)^{3}=-1225043 \](viii) \(\displaystyle (-299)^{3}=-\left(299^{3}\right)\), and \(\displaystyle 299^{3}=(300-1)^{3}\): \[=27000000-3(90000)(1)+3(300)(1)-1=27000000-270000+900-1=26730899. \] \[(-299)^{3}=-26730899 \]Answers: (i) \(\displaystyle 357\); (ii) \(\displaystyle 9984\); (iii) \(\displaystyle 384\); (iv) \(\displaystyle 3176523\); (v) \(\displaystyle 7880599\); (vi) \(\displaystyle 2048383\); (vii) \(\displaystyle -1225043\); (viii) \(\displaystyle -26730899\).

Identify the pattern before you factor. Two terms that are both cubes point to \(\displaystyle a^{3}\pm b^{3}\); two terms that are both squares point to \(\displaystyle a^{2}-b^{2}\); three terms with two squares point to \(\displaystyle (a\pm b)^{2}\); four terms with two cubes point to \(\displaystyle (a\pm b)^{3}\); and four terms of the form \(\displaystyle a^{3}+b^{3}+c^{3}-3abc\) point to \(\displaystyle (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)\).(i) \(\displaystyle 4y^{2}+1+\frac{1}{16y^{2}}\). Outer squares \(\displaystyle (2y)^{2}\) and \(\displaystyle \left(\frac{1}{4y}\right)^{2}\); check \(\displaystyle 2(2y)\left(\frac{1}{4y}\right)=1\), the middle term. \[=\left(2y+\tfrac{1}{4y}\right)^{2} \](ii) \(\displaystyle 9m^{2}-\frac{1}{25n^{2}}=(3m)^{2}-\left(\frac{1}{5n}\right)^{2}\), a difference of squares. \[=\left(3m-\tfrac{1}{5n}\right)\left(3m+\tfrac{1}{5n}\right) \](iii) \(\displaystyle 27b^{3}-\frac{1}{64b^{3}}=(3b)^{3}-\left(\frac{1}{4b}\right)^{3}\). Use \(\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})\), where \(\displaystyle ab=3b\cdot\frac{1}{4b}=\frac{3}{4}\). \[=\left(3b-\tfrac{1}{4b}\right)\left(9b^{2}+\tfrac{3}{4}+\tfrac{1}{16b^{2}}\right) \](iv) \(\displaystyle x^{2}+\frac{5x}{6}+\frac{1}{6}\). Multiply and divide by \(\displaystyle 6\) to clear fractions: \(\displaystyle \frac{1}{6}\left(6x^{2}+5x+1\right)\). Split \(\displaystyle 5x\) using \(\displaystyle 2\) and \(\displaystyle 3\) (product \(\displaystyle 6\times 1=6\), sum \(\displaystyle 5\)): \[6x^{2}+5x+1=6x^{2}+3x+2x+1=3x(2x+1)+(2x+1)=(2x+1)(3x+1). \] \[x^{2}+\tfrac{5x}{6}+\tfrac{1}{6}=\tfrac{1}{6}(2x+1)(3x+1)=\left(x+\tfrac{1}{2}\right)\left(x+\tfrac{1}{3}\right) \](v) \(\displaystyle 27u^{3}-\frac{1}{125}-\frac{27u^{2}}{5}+\frac{9u}{25}\). Four terms with \(\displaystyle (3u)^{3}\) and \(\displaystyle \left(\frac{1}{5}\right)^{3}\) suggests \(\displaystyle (a-b)^{3}\) with \(\displaystyle a=3u\), \(\displaystyle b=\frac{1}{5}\). Check the two middle terms: \(\displaystyle 3a^{2}b=3(9u^{2})\left(\frac{1}{5}\right)=\frac{27u^{2}}{5}\) (subtracted ✔) and \(\displaystyle 3ab^{2}=3(3u)\left(\frac{1}{25}\right)=\frac{9u}{25}\) (added ✔). \[=\left(3u-\tfrac{1}{5}\right)^{3} \](vi) \(\displaystyle 64y^{3}+\frac{1}{125}z^{3}=(4y)^{3}+\left(\frac{z}{5}\right)^{3}\). Use \(\displaystyle a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})\). \[=\left(4y+\tfrac{z}{5}\right)\left(16y^{2}-\tfrac{4yz}{5}+\tfrac{z^{2}}{25}\right) \](vii) \(\displaystyle p^{3}+27q^{3}+r^{3}-9pqr\). Write \(\displaystyle 27q^{3}=(3q)^{3}\); then with \(\displaystyle a=p\), \(\displaystyle b=3q\), \(\displaystyle c=r\) we get \(\displaystyle 3abc=3(p)(3q)(r)=9pqr\), so this is exactly \(\displaystyle a^{3}+b^{3}+c^{3}-3abc\). \[=(p+3q+r)\left(p^{2}+9q^{2}+r^{2}-3pq-3qr-rp\right) \](viii) \(\displaystyle 9m^{2}-12m+4\): squares \(\displaystyle (3m)^{2}\) and \(\displaystyle 2^{2}\), and \(\displaystyle 2(3m)(2)=12m\). \[=(3m-2)^{2} \](ix) \(\displaystyle 9x^{3}-\frac{8}{3}y^{3}+\frac{z^{3}}{3}+6xyz\). None of \(\displaystyle 9x^{3}\), \(\displaystyle \frac{8}{3}y^{3}\), \(\displaystyle \frac{z^{3}}{3}\) is a clean cube, so take out \(\displaystyle \frac{1}{3}\) first: \[=\tfrac{1}{3}\left(27x^{3}-8y^{3}+z^{3}+18xyz\right). \] Now \(\displaystyle 27x^{3}=(3x)^{3}\), \(\displaystyle -8y^{3}=(-2y)^{3}\), \(\displaystyle z^{3}=z^{3}\), and with \(\displaystyle a=3x\), \(\displaystyle b=-2y\), \(\displaystyle c=z\) we get \(\displaystyle -3abc=-3(3x)(-2y)(z)=+18xyz\) ✔. So the bracket is \(\displaystyle a^{3}+b^{3}+c^{3}-3abc\) and \[a^{2}+b^{2}+c^{2}-ab-bc-ca=9x^{2}+4y^{2}+z^{2}+6xy+2yz-3xz. \] \[9x^{3}-\tfrac{8}{3}y^{3}+\tfrac{z^{3}}{3}+6xyz=\tfrac{1}{3}(3x-2y+z)\left(9x^{2}+4y^{2}+z^{2}+6xy+2yz-3xz\right) \](x) \(\displaystyle 4x^{2}+9y^{2}+36z^{2}+12xz+36yz+24xy\). This one does not factor as printed, and I believe the printed cross terms are swapped. Here is the check. Six terms with the squares \(\displaystyle (2x)^{2}\), \(\displaystyle (3y)^{2}\), \(\displaystyle (6z)^{2}\) invite \(\displaystyle (2x+3y+6z)^{2}\), whose cross terms are forced to be \[2(2x)(3y)=12xy,\qquad 2(3y)(6z)=36yz,\qquad 2(2x)(6z)=24xz. \] The printed expression has \(\displaystyle 36yz\) ✔ but \(\displaystyle 24xy\) and \(\displaystyle 12xz\) — the coefficients \(\displaystyle 24\) and \(\displaystyle 12\) are attached to the wrong pairs. Changing the signs of \(\displaystyle 2x\), \(\displaystyle 3y\) or \(\displaystyle 6z\) cannot fix this, because it changes signs, not sizes. A quick numerical check confirms the mismatch: at \(\displaystyle x=y=1\), \(\displaystyle z=0\) the printed expression gives \(\displaystyle 4+9+24=37\), while \(\displaystyle (2x+3y+6z)^{2}=(2+3)^{2}=25\). In fact the printed expression cannot be written as a product of two linear expressions at all, so it is not factorable by any identity in this chapter.Almost certainly the intended expression is \(\displaystyle 4x^{2}+9y^{2}+36z^{2}+12xy+36yz+24xz\), and then \[4x^{2}+9y^{2}+36z^{2}+12xy+36yz+24xz=(2x+3y+6z)^{2}. \](xi) \(\displaystyle 27u^{3}-\frac{1}{216}-\frac{9u^{2}}{2}+\frac{u}{4}\). Here \(\displaystyle (3u)^{3}=27u^{3}\) and \(\displaystyle \left(\frac{1}{6}\right)^{3}=\frac{1}{216}\), so try \(\displaystyle (a-b)^{3}\) with \(\displaystyle a=3u\), \(\displaystyle b=\frac{1}{6}\). Check: \(\displaystyle 3a^{2}b=3(9u^{2})\left(\frac{1}{6}\right)=\frac{9u^{2}}{2}\) (subtracted ✔) and \(\displaystyle 3ab^{2}=3(3u)\left(\frac{1}{36}\right)=\frac{u}{4}\) (added ✔). \[=\left(3u-\tfrac{1}{6}\right)^{3} \]Answers: (i) \(\displaystyle \left(2y+\frac{1}{4y}\right)^{2}\); (ii) \(\displaystyle \left(3m-\frac{1}{5n}\right)\left(3m+\frac{1}{5n}\right)\); (iii) \(\displaystyle \left(3b-\frac{1}{4b}\right)\left(9b^{2}+\frac{3}{4}+\frac{1}{16b^{2}}\right)\); (iv) \(\displaystyle \left(x+\frac{1}{2}\right)\left(x+\frac{1}{3}\right)\); (v) \(\displaystyle \left(3u-\frac{1}{5}\right)^{3}\); (vi) \(\displaystyle \left(4y+\frac{z}{5}\right)\left(16y^{2}-\frac{4yz}{5}+\frac{z^{2}}{25}\right)\); (vii) \(\displaystyle (p+3q+r)\left(p^{2}+9q^{2}+r^{2}-3pq-3qr-rp\right)\); (viii) \(\displaystyle (3m-2)^{2}\); (ix) \(\displaystyle \frac{1}{3}(3x-2y+z)\left(9x^{2}+4y^{2}+z^{2}+6xy+2yz-3xz\right)\); (x) not factorable as printed — with the \(\displaystyle xy\) and \(\displaystyle xz\) coefficients in their intended places it is \(\displaystyle (2x+3y+6z)^{2}\); (xi) \(\displaystyle \left(3u-\frac{1}{6}\right)^{3}\).

Factor top and bottom, then cancel common factors. Cancelling is only allowed for whole factors, so the first job is always to turn each sum into a product.(i) \(\displaystyle \dfrac{4x^{2}+4x+1}{4x^{2}-1}\). The numerator has squares \(\displaystyle (2x)^{2}\) and \(\displaystyle 1^{2}\) with \(\displaystyle 2(2x)(1)=4x\), so it is \(\displaystyle (2x+1)^{2}\). The denominator is a difference of squares, \(\displaystyle (2x-1)(2x+1)\). \[\frac{(2x+1)^{2}}{(2x-1)(2x+1)}=\frac{2x+1}{2x-1} \](ii) \(\displaystyle \dfrac{9\left(3a^{3}-24b^{3}\right)}{9a^{2}-36b^{2}}\). Numerator: \(\displaystyle 9\cdot 3\left(a^{3}-8b^{3}\right)=27\left(a^{3}-(2b)^{3}\right)=27(a-2b)\left(a^{2}+2ab+4b^{2}\right)\). Denominator: \(\displaystyle 9\left(a^{2}-4b^{2}\right)=9(a-2b)(a+2b)\). \[\frac{27(a-2b)\left(a^{2}+2ab+4b^{2}\right)}{9(a-2b)(a+2b)}=\frac{3\left(a^{2}+2ab+4b^{2}\right)}{a+2b} \](iii) \(\displaystyle \dfrac{s^{3}+125t^{3}}{s^{2}-2st-35t^{2}}\). Numerator: \(\displaystyle s^{3}+(5t)^{3}=(s+5t)\left(s^{2}-5st+25t^{2}\right)\). Denominator: two numbers with product \(\displaystyle -35\) and sum \(\displaystyle -2\) are \(\displaystyle -7\) and \(\displaystyle +5\), so it is \(\displaystyle (s-7t)(s+5t)\). \[\frac{(s+5t)\left(s^{2}-5st+25t^{2}\right)}{(s-7t)(s+5t)}=\frac{s^{2}-5st+25t^{2}}{s-7t} \]Answers: (i) \(\displaystyle \frac{2x+1}{2x-1}\); (ii) \(\displaystyle \frac{3\left(a^{2}+2ab+4b^{2}\right)}{a+2b}\); (iii) \(\displaystyle \frac{s^{2}-5st+25t^{2}}{s-7t}\).

Area of a rectangle \(\displaystyle =\) length \(\displaystyle \times\) breadth, so factoring the area gives possible side lengths. Every way of writing the area as a product of two expressions is a legitimate answer — which is why the question says "possible expressions". The natural answer is the factorisation into the simplest pieces.(i) Area \(\displaystyle =25a^{2}-30ab+9b^{2}\). The outer terms are \(\displaystyle (5a)^{2}\) and \(\displaystyle (3b)^{2}\), and \(\displaystyle 2(5a)(3b)=30ab\) matches the middle term, which is negative: \[25a^{2}-30ab+9b^{2}=(5a-3b)^{2}=(5a-3b)\times(5a-3b). \] So one possible rectangle has length \(\displaystyle =5a-3b\) and breadth \(\displaystyle =5a-3b\) — the two sides are equal, so this rectangle is in fact a square.(ii) Area \(\displaystyle =36s^{2}-49t^{2}=(6s)^{2}-(7t)^{2}\), a difference of squares: \[36s^{2}-49t^{2}=(6s+7t)(6s-7t). \] So length \(\displaystyle =6s+7t\) and breadth \(\displaystyle =6s-7t\) (taking the longer factor as the length).Other answers are equally valid. Nothing forces this particular split — for example \(\displaystyle (5a-3b)\) and \(\displaystyle (5a-3b)\) could be replaced by \(\displaystyle 2(5a-3b)\) and \(\displaystyle \frac{1}{2}(5a-3b)\), and in (ii) by \(\displaystyle 3(6s+7t)\) and \(\displaystyle \frac{1}{3}(6s-7t)\). Any pair of expressions whose product is the given area describes a rectangle of that area; the factorisations above are simply the tidiest ones.Answers: (i) length \(\displaystyle =5a-3b\), breadth \(\displaystyle =5a-3b\); (ii) length \(\displaystyle =6s+7t\), breadth \(\displaystyle =6s-7t\). (Other factor pairs with the same product are also correct.)

Volume of a cuboid \(\displaystyle =\) length \(\displaystyle \times\) breadth \(\displaystyle \times\) height, so split the volume into three factors. As in the previous question the answer is not unique — any three expressions with the given product will do — so the aim is a tidy factorisation.(i) Volume \(\displaystyle =6a^{2}-24b^{2}\). Take out the common factor \(\displaystyle 6\), then use the difference of squares: \[6a^{2}-24b^{2}=6\left(a^{2}-4b^{2}\right)=6\left(a^{2}-(2b)^{2}\right)=6(a-2b)(a+2b). \] That is a product of three factors, so one possible cuboid has length \(\displaystyle =a+2b\), breadth \(\displaystyle =a-2b\), height \(\displaystyle =6\).(ii) Volume \(\displaystyle =3ps^{2}-15ps+12p\). Every term has \(\displaystyle 3p\) in it, so take that out first, then split the middle term of the quadratic (two numbers with product \(\displaystyle 4\) and sum \(\displaystyle -5\): \(\displaystyle -1\) and \(\displaystyle -4\)): \[3ps^{2}-15ps+12p=3p\left(s^{2}-5s+4\right)=3p(s-1)(s-4). \] So one possible cuboid has length \(\displaystyle =s-1\), breadth \(\displaystyle =s-4\), height \(\displaystyle =3p\).Other answers are equally valid. In (i) the \(\displaystyle 6\) could be split as \(\displaystyle 2\) and \(\displaystyle 3\) and shared out differently — for instance length \(\displaystyle 2(a+2b)\), breadth \(\displaystyle 3(a-2b)\), height \(\displaystyle 1\) — and in (ii) the \(\displaystyle 3p\) could be split as \(\displaystyle 3\) and \(\displaystyle p\), giving height \(\displaystyle 3\), and \(\displaystyle p(s-1)\) as one of the other edges. Any three expressions whose product is the given volume answer the question; the versions above are the simplest.Answers: (i) length \(\displaystyle =a+2b\), breadth \(\displaystyle =a-2b\), height \(\displaystyle =6\); (ii) length \(\displaystyle =s-1\), breadth \(\displaystyle =s-4\), height \(\displaystyle =3p\). (Other factor triples with the same product are also correct.)

Area of the big square minus the area of the playground.The path is built around the playground, so the playground keeps its full side of $\displaystyle 40$ m and the path is a border of width \(\displaystyle s\) m lying outside it.Walk across the whole figure along one direction: you cross the path (\(\displaystyle s\) m), then the playground ($\displaystyle 40$ m), then the path again (\(\displaystyle s\) m). So the outer boundary is itself a square, of side\[40 + s + s \;=\; 40 + 2s \ \text{metres}. \]The path is what is left when the playground is removed from this big square:\[\text{Area of path} \;=\; (40+2s)^2 - 40^2. \]Expand \(\displaystyle (40+2s)^2\) with the identity \(\displaystyle (a+b)^2=a^2+2ab+b^2\), taking \(\displaystyle a=40,\ b=2s\):\[(40+2s)^2 = 40^2 + 2\cdot 40\cdot 2s + (2s)^2 = 1600 + 160s + 4s^2. \]\[\text{Area of path} = 1600 + 160s + 4s^2 - 1600 = 4s^2 + 160s. \]Check $\displaystyle 1$ (difference of squares). Using \(\displaystyle a^2-b^2=(a+b)(a-b)\) with \(\displaystyle a=40+2s,\ b=40\): \[(40+2s)^2-40^2 = \big((40+2s)+40\big)\big((40+2s)-40\big) = (80+2s)(2s) = 160s + 4s^2, \] the same expression.Check $\displaystyle 2$ (cut the path into pieces). The path is four strips of size \(\displaystyle 40 \times s\) along the four sides, plus four small squares of side \(\displaystyle s\) at the corners: \[4(40s) + 4(s^2) = 160s + 4s^2. \] Again the same. For example, with \(\displaystyle s=1\) all three methods give \(\displaystyle 164\) square metres.Area of the path \(\displaystyle = 4s^2 + 160s = 4s(s+40)\) square metres.

Clear the fraction to get a quadratic, then split the middle term.Let the number be \(\displaystyle x\). A number has a reciprocal only if it is not zero, so \(\displaystyle x \neq 0\), and the condition is\[x + \frac{1}{x} = \frac{10}{3}. \]Multiply both sides by \(\displaystyle 3x\). This is allowed because \(\displaystyle x \neq 0\), so \(\displaystyle 3x \neq 0\) and we are not multiplying by zero:\[3x\cdot x + 3x\cdot\frac{1}{x} = 3x \cdot \frac{10}{3} \quad\Longrightarrow\quad 3x^2 + 3 = 10x. \]Bring everything to one side:\[3x^2 - 10x + 3 = 0. \]Now split the middle term. We need two numbers whose product is \(\displaystyle 3 \times 3 = 9\) and whose sum is \(\displaystyle -10\); these are \(\displaystyle -9\) and \(\displaystyle -1\):\[3x^2 - 9x - x + 3 = 3x(x-3) - 1(x-3) = (x-3)(3x-1). \]So \(\displaystyle (x-3)(3x-1)=0\), which gives \(\displaystyle x = 3\) or \(\displaystyle x = \dfrac{1}{3}\).Verification. \[3 + \frac{1}{3} = \frac{9+1}{3} = \frac{10}{3} \qquad\text{and}\qquad \frac{1}{3} + 3 = \frac{10}{3}. \] Both work.It is no accident that the two answers are reciprocals of each other: the equation \(\displaystyle x+\frac1x=\frac{10}{3}\) reads exactly the same if \(\displaystyle x\) is replaced by \(\displaystyle \frac1x\), so whenever a number works, its reciprocal works too.The number is \(\displaystyle 3\) (or, equally, its reciprocal \(\displaystyle \frac{1}{3}\)).

Factorise the area; the given width has to be one of the factors.For a rectangle, \(\displaystyle \text{area} = \text{length} \times \text{width}\). So if we can write the area \(\displaystyle 2x^2+7x+3\) as a product of two expressions, one of which is the width \(\displaystyle 2x+1\), the other one must be the length.Factorise \(\displaystyle 2x^2+7x+3\) by splitting the middle term. We need two numbers whose product is \(\displaystyle 2 \times 3 = 6\) and whose sum is \(\displaystyle 7\); these are \(\displaystyle 6\) and \(\displaystyle 1\):\[2x^2+7x+3 = 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (x+3)(2x+1). \]So\[(2x+1)\times \text{length} = (2x+1)(x+3), \]and the length is \(\displaystyle x+3\).Verification — multiply back: \[(2x+1)(x+3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3, \] which is the given area.Length \(\displaystyle = x+3\) hastas.

Factor Theorem, used twice.The Factor Theorem says: \(\displaystyle (x-a)\) is a factor of a polynomial \(\displaystyle f(x)\) exactly when \(\displaystyle f(a)=0\).Let \(\displaystyle f(x) = px^2 + 5x + r\).Because \(\displaystyle x-2\) is a factor, \(\displaystyle f(2)=0\): \[p(2)^2 + 5(2) + r = 0 \quad\Longrightarrow\quad 4p + 10 + r = 0 \quad\Longrightarrow\quad 4p + r = -10. \tag{1}\]Because \(\displaystyle x-\frac{1}{2}\) is a factor, \(\displaystyle f\!\left(\frac12\right)=0\): \[p\left(\frac12\right)^2 + 5\left(\frac12\right) + r = 0 \quad\Longrightarrow\quad \frac{p}{4} + \frac{5}{2} + r = 0. \] Multiply through by \(\displaystyle 4\) to clear the fractions: \[p + 10 + 4r = 0 \quad\Longrightarrow\quad p + 4r = -10. \tag{2}\]Compare ($\displaystyle 1$) and ($\displaystyle 2$). Both left-hand sides equal \(\displaystyle -10\), so they equal each other: \[4p + r = p + 4r. \] Take \(\displaystyle p\) and \(\displaystyle r\) to opposite sides: \[4p - p = 4r - r \quad\Longrightarrow\quad 3p = 3r \quad\Longrightarrow\quad p = r. \]Hence \(\displaystyle p=r\), as required.A look at the actual polynomial (not required, but reassuring): putting \(\displaystyle r=p\) into ($\displaystyle 1$) gives \(\displaystyle 5p=-10\), so \(\displaystyle p=r=-2\) and \[f(x) = -2x^2 + 5x - 2 = -(2x^2-5x+2) = -(2x-1)(x-2), \] which really does have both \(\displaystyle x-2\) and \(\displaystyle x-\frac12\) (inside \(\displaystyle 2x-1\)) as factors.

The identity \(\displaystyle a^3+b^3+c^3-3abc = (a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)\).The identity turns the required expression into a product, and we already know one of the factors: \(\displaystyle a+b+c=5\). So we only need the value of \(\displaystyle a^2+b^2+c^2-ab-bc-ca\).Step $\displaystyle 1$ — find \(\displaystyle a^2+b^2+c^2\). Use \[(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca). \] Substituting \(\displaystyle a+b+c=5\) and \(\displaystyle ab+bc+ca=10\): \[5^2 = a^2+b^2+c^2 + 2(10) \quad\Longrightarrow\quad 25 = a^2+b^2+c^2 + 20 \quad\Longrightarrow\quad a^2+b^2+c^2 = 5. \]Step $\displaystyle 2$ — form the second factor. \[a^2+b^2+c^2-ab-bc-ca = 5 - 10 = -5. \]Step $\displaystyle 3$ — multiply. \[a^3+b^3+c^3-3abc = (a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right) = 5 \times (-5) = -25. \]Hence \(\displaystyle a^3+b^3+c^3-3abc = -25\).A remark for the careful reader. The identity can also be written as \[a^3+b^3+c^3-3abc = \tfrac{1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right], \] and here \(\displaystyle (a-b)^2+(b-c)^2+(c-a)^2 = 2(a^2+b^2+c^2)-2(ab+bc+ca) = 10-20 = -10\), giving \(\displaystyle \tfrac12 \times 5 \times (-10) = -25\) again. But a sum of squares of real numbers can never be negative, so no real numbers \(\displaystyle a,b,c\) actually satisfy both of the given conditions at once. The proof above is still exactly the proof the question asks for — every step follows from the two given equations and a correct identity — it is just worth knowing that the given data could not come from real measurements.

Factorise into three consecutive numbers.Step $\displaystyle 1$ — factorise. \[n^3 - n = n(n^2-1) = n(n-1)(n+1) = (n-1)\,n\,(n+1), \] using the difference of squares \(\displaystyle n^2-1=(n-1)(n+1)\). So \(\displaystyle n^3-n\) is the product of the three consecutive integers \(\displaystyle n-1,\ n,\ n+1\).Step $\displaystyle 2$ — why the product is divisible by 2. Out of any two consecutive integers, one is even. So at least one of \(\displaystyle n\) and \(\displaystyle n+1\) is even, and a product containing an even factor is even. Hence \(\displaystyle 2\) divides \(\displaystyle n^3-n\).Step $\displaystyle 3$ — why the product is divisible by 3. Divide \(\displaystyle n\) by \(\displaystyle 3\); the remainder is \(\displaystyle 0\), \(\displaystyle 1\) or \(\displaystyle 2\), so \(\displaystyle n\) has one of the forms \(\displaystyle 3k,\ 3k+1,\ 3k+2\).

Form of \(\displaystyle n\)	A multiple of $\displaystyle 3$ among the three	Why
\(\displaystyle n=3k\)	\(\displaystyle n\)	\(\displaystyle n=3k\)
\(\displaystyle n=3k+1\)	\(\displaystyle n-1\)	\(\displaystyle n-1=3k\)
\(\displaystyle n=3k+2\)	\(\displaystyle n+1\)	\(\displaystyle n+1=3k+3=3(k+1)\)

In every case one of \(\displaystyle n-1,\ n,\ n+1\) is a multiple of \(\displaystyle 3\), so \(\displaystyle 3\) divides the product.Step $\displaystyle 4$ — put the two together. The number \(\displaystyle n^3-n\) is divisible by \(\displaystyle 2\) and by \(\displaystyle 3\). Since \(\displaystyle 2\) and \(\displaystyle 3\) have no common factor other than \(\displaystyle 1\), their product \(\displaystyle 2\times 3=6\) must also divide it. (This last step needs the "no common factor" part: being divisible by \(\displaystyle 2\) and by \(\displaystyle 4\) would not force divisibility by \(\displaystyle 8\), because \(\displaystyle 2\) and \(\displaystyle 4\) share the factor \(\displaystyle 2\).)Spot checks. \(\displaystyle n=1:\ 1-1=0=6\times 0\). \(\displaystyle n=2:\ 8-2=6=6\times1\). \(\displaystyle n=5:\ 125-5=120=6\times 20\). \(\displaystyle n=10:\ 1000-10=990=6\times165\). (A computer check of every \(\displaystyle n\) up to \(\displaystyle 2000\) also leaves remainder \(\displaystyle 0\) each time.)Hence \(\displaystyle n^3-n = (n-1)n(n+1)\) is divisible by $\displaystyle 6$ for every natural number \(\displaystyle n\).

Recognise each expression as \(\displaystyle a^3+b^3+c^3-3abc\), which factorises as \(\displaystyle (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).The point of both parts is the same: if the expression can be written as \(\displaystyle a^3+b^3+c^3-3abc\), it carries \(\displaystyle (a+b+c)\) as a factor — and in each part the given condition makes \(\displaystyle a+b+c\) equal to zero.(i) \(\displaystyle x^3+y^3-12xy+64\), when \(\displaystyle x+y=-4\).Match the pieces. Take \(\displaystyle a=x,\ b=y,\ c=4\). Then \[a^3+b^3+c^3 = x^3+y^3+4^3 = x^3+y^3+64 \quad\checkmark \] \[-3abc = -3\cdot x\cdot y\cdot 4 = -12xy \quad\checkmark \] so the expression is exactly \(\displaystyle a^3+b^3+c^3-3abc\) for these \(\displaystyle a,b,c\). Therefore \[x^3+y^3+64-12xy = (x+y+4)\left(x^2+y^2+16-xy-4y-4x\right). \]Now use the condition. Since \(\displaystyle x+y=-4\), \[x+y+4 = -4+4 = 0, \] so the first factor is \(\displaystyle 0\) and the whole product is \(\displaystyle 0\), whatever the second factor happens to be.Value \(\displaystyle = 0\).Quick check: take \(\displaystyle x=1,\ y=-5\) (so \(\displaystyle x+y=-4\)): \(\displaystyle 1 + (-125) - 12(1)(-5) + 64 = 1-125+60+64 = 0.\ \checkmark\)(ii) \(\displaystyle x^3-8y^3-36xy-216\), when \(\displaystyle x=2y+6\).Here the minus signs are absorbed into the letters. Take \(\displaystyle a=x,\ b=-2y,\ c=-6\). Then \[a^3 = x^3, \qquad b^3 = (-2y)^3 = -8y^3, \qquad c^3 = (-6)^3 = -216 \quad\checkmark \] \[-3abc = -3\cdot x\cdot(-2y)\cdot(-6) = -3\cdot 12xy = -36xy \quad\checkmark \] so again the expression is \(\displaystyle a^3+b^3+c^3-3abc\), and \[x^3-8y^3-216-36xy = (x-2y-6)\left(x^2+4y^2+36+2xy+6x-12y\right). \]Now use the condition. Since \(\displaystyle x=2y+6\), \[x-2y-6 = (2y+6)-2y-6 = 0, \] so the first factor is \(\displaystyle 0\) and the product is \(\displaystyle 0\).Value \(\displaystyle = 0\).Quick check: take \(\displaystyle y=1\), so \(\displaystyle x=8\): \(\displaystyle 512 - 8(1) - 36(8)(1) - 216 = 512-8-288-216 = 0.\ \checkmark\)Both values are \(\displaystyle 0\).