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NCERT Exemplar · Class 9 Mathematics Polynomials

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EXERCISE 2.3 31–40 (part 7 of 8)

  1. Exercise 31

    Expand the following: (i) (3a−2b)3\displaystyle (3 a-2 b)^{3} (ii) (1x+y3)3\displaystyle \left(\frac{1}{x}+\frac{y}{3}\right)^{3} (iii) (4−13x)3\displaystyle \left(4-\frac{1}{3 x}\right)^{3}

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    NCERT’s answer
    (i)
    $\displaystyle 27 a^{3}-54 a^{2} b+36 a b^{2}-8 b^{3}$ (ii) $\displaystyle \frac{1}{x^{3}}+\frac{y}{x^{2}}+\frac{y^{2}}{3 x}+\frac{y^{3}}{27}$ (iii) $\displaystyle 64-\frac{16}{x}+\frac{4}{3 x^{2}}-\frac{1}{27 x^{3}}$
    \[(p-q)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} \] (i) \(\displaystyle p=3a,\ q=2b\) \[(3a-2b)^{3}=(3a)^{3}-3(3a)^{2}(2b)+3(3a)(2b)^{2}-(2b)^{3} \] \[=27a^{3}-54a^{2}b+36ab^{2}-8b^{3} \] \[(p+q)^{3}=p^{3}+3p^{2}q+3pq^{2}+q^{3} \] (ii) \(\displaystyle p=\dfrac1x,\ q=\dfrac y3\) \[\left(\frac1x+\frac y3\right)^{3}=\left(\frac1x\right)^{3}+3\left(\frac1x\right)^{2}\left(\frac y3\right)+3\left(\frac1x\right)\left(\frac y3\right)^{2}+\left(\frac y3\right)^{3} \] \[=\frac{1}{x^{3}}+\frac{y}{x^{2}}+\frac{y^{2}}{3x}+\frac{y^{3}}{27} \] (iii) \(\displaystyle p=4,\ q=\dfrac{1}{3x}\), a difference cube again \[\left(4-\frac{1}{3x}\right)^{3}=4^{3}-3(4)^{2}\left(\frac{1}{3x}\right)+3(4)\left(\frac{1}{3x}\right)^{2}-\left(\frac{1}{3x}\right)^{3} \] \[=64-\frac{16}{x}+\frac{4}{3x^{2}}-\frac{1}{27x^{3}} \] Answer: (i) \(\displaystyle 27a^{3}-54a^{2}b+36ab^{2}-8b^{3}\) (ii) \(\displaystyle \dfrac{1}{x^{3}}+\dfrac{y}{x^{2}}+\dfrac{y^{2}}{3x}+\dfrac{y^{3}}{27}\) (iii) \(\displaystyle 64-\dfrac{16}{x}+\dfrac{4}{3x^{2}}-\dfrac{1}{27x^{3}}\)

  2. Exercise 32

    Factorise the following: (i) 1−64a3−12a+48a2\displaystyle 1-64 a^{3}-12 a+48 a^{2} (ii) 8p3+125p2+625p+1125\displaystyle 8 p^{3}+\frac{12}{5} p^{2}+\frac{6}{25} p+\frac{1}{125}

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    NCERT’s answer
    (i)
    $\displaystyle (1-4 a)(1-4 a)(1-4 a)$ (ii) $\displaystyle \left(2 p+\frac{1}{5}\right)\left(2 p+\frac{1}{5}\right)\left(2 p+\frac{1}{5}\right)$
    \[(1-4a)^{3}=1-3(4a)+3(4a)^{2}-(4a)^{3}=1-12a+48a^{2}-64a^{3} \] \[\text{(i) } 1-64a^{3}-12a+48a^{2}=1-12a+48a^{2}-64a^{3}=(1-4a)^{3} \] \[\left(2p+\frac15\right)^{3}=8p^{3}+3(2p)^{2}\!\left(\frac15\right)+3(2p)\!\left(\frac15\right)^{2}+\frac{1}{125}=8p^{3}+\frac{12}{5}p^{2}+\frac{6}{25}p+\frac{1}{125} \] \[\text{(ii) } 8p^{3}+\frac{12}{5}p^{2}+\frac{6}{25}p+\frac{1}{125}=\left(2p+\frac15\right)^{3} \]Answer: (i) \(\displaystyle (1-4a)^{3}\) (ii) \(\displaystyle \left(2p+\frac15\right)^{3}\)

  3. Exercise 33

    Find the following products : (i) (x2+2y)(x24−xy+4y2)\displaystyle \left(\frac{x}{2}+2 y\right)\left(\frac{x^{2}}{4}-x y+4 y^{2}\right) (ii) (x2−1)(x4+x2+1)\displaystyle \left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)

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    NCERT’s answer
    (i)
    $\displaystyle \frac{x^{3}}{8}+8 y^{3}$ (ii) $\displaystyle x^{6}-1$
    (i) \[\left(\frac{x}{2}\right)^{2}-\left(\frac{x}{2}\right)(2y)+(2y)^{2}=\frac{x^{2}}{4}-xy+4y^{2} \] \[\left(\frac{x}{2}+2y\right)\left(\frac{x^{2}}{4}-xy+4y^{2}\right)=\left(\frac{x}{2}\right)^{3}+(2y)^{3} \quad \text{(sum of cubes)} \] \[=\frac{x^{3}}{8}+8y^{3} \] (ii) \[(x^{2}-1)(x^{4}+x^{2}+1)=(x^{2})^{3}-1^{3} \quad \text{(difference of cubes)} \] \[=x^{6}-1 \] Answer: (i) \(\displaystyle \dfrac{x^{3}}{8}+8y^{3}\) (ii) \(\displaystyle x^{6}-1\)

  4. Exercise 34

    Factorise : (i) 1+64x3\displaystyle 1+64 x^{3} (ii) a3−22b3\displaystyle a^{3}-2 \sqrt{2} b^{3}

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    NCERT’s answer
    (i)
    $\displaystyle (1+4 x)\left(1-4 x+16 x^{2}\right)$ (ii) $\displaystyle (a-\sqrt{2} b)\left(a^{2}+\sqrt{2} a b+2 b^{2}\right)$
    (i) \[1+64x^{3}=1^{3}+(4x)^{3} \] \[=(1+4x)\big(1-4x+16x^{2}\big) \quad \text{(sum of cubes)} \] (ii) \[2\sqrt2\,b^{3}=(\sqrt2\,b)^{3} \] \[a^{3}-(\sqrt2\,b)^{3}=(a-\sqrt2\,b)\big(a^{2}+\sqrt2\,ab+2b^{2}\big) \quad \text{(difference of cubes)} \] Answer: (i) \(\displaystyle (1+4x)(1-4x+16x^{2})\) (ii) \(\displaystyle (a-\sqrt2\,b)(a^{2}+\sqrt2\,ab+2b^{2})\)

  5. Exercise 35

    Find the following product: (2x−y+3z)(4x2+y2+9z2+2xy+3yz−6xz)(2 x-y+3 z)\left(4 x^{2}+y^{2}+9 z^{2}+2 x y+3 y z-6 x z\right)

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    NCERT’s answer
    $\displaystyle 8 x^{3}-y^{3}+27 z^{3}+18 x y z$
    \[a=2x,\ b=-y,\ c=3z \] \[a^{2}+b^{2}+c^{2}-ab-bc-ca=4x^{2}+y^{2}+9z^{2}+2xy+3yz-6xz \] \[(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)=a^{3}+b^{3}+c^{3}-3abc \] \[(2x-y+3z)(4x^{2}+y^{2}+9z^{2}+2xy+3yz-6xz)=(2x)^{3}+(-y)^{3}+(3z)^{3}-3(2x)(-y)(3z) \] \[=8x^{3}-y^{3}+27z^{3}+18xyz \] Answer: \(\displaystyle 8x^{3}-y^{3}+27z^{3}+18xyz\)

  6. Exercise 36

    Factorise : (i) a3−8b3−64c3−24abc\displaystyle a^{3}-8 b^{3}-64 c^{3}-24 a b c (ii) 22a3+8b3−27c3+182abc\displaystyle 2 \sqrt{2} a^{3}+8 b^{3}-27 c^{3}+18 \sqrt{2} a b c.

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    NCERT’s answer
    (i)
    $\displaystyle (a-2 b-4 c)\left(a^{2}+4 b^{2}+16 c^{2}+2 a b-8 b c+4 a c\right)$ (ii) $\displaystyle (\sqrt{2} a+2 b-3 c)\left(2 a^{2}+4 b^{2}+9 c^{2}-2 \sqrt{2} a b+6 b c+3 \sqrt{2} a c\right)$
    (i) \[a^{3}-8b^{3}-64c^{3}-24abc=a^{3}+(-2b)^{3}+(-4c)^{3}-3a(-2b)(-4c) \] \[=\big(a-2b-4c\big)\big(a^{2}+4b^{2}+16c^{2}+2ab-8bc+4ac\big) \quad \text{(three-cube identity)} \] (ii) \[2\sqrt2\,a^{3}+8b^{3}-27c^{3}+18\sqrt2\,abc=(\sqrt2\,a)^{3}+(2b)^{3}+(-3c)^{3}-3(\sqrt2\,a)(2b)(-3c) \] \[=\big(\sqrt2\,a+2b-3c\big)\big(2a^{2}+4b^{2}+9c^{2}-2\sqrt2\,ab+6bc+3\sqrt2\,ac\big) \quad \text{(three-cube identity)} \] Answer: (i) \(\displaystyle (a-2b-4c)(a^{2}+4b^{2}+16c^{2}+2ab-8bc+4ac)\) (ii) \(\displaystyle (\sqrt2\,a+2b-3c)(2a^{2}+4b^{2}+9c^{2}-2\sqrt2\,ab+6bc+3\sqrt2\,ac)\)

  7. Exercise 37

    Without actually calculating the cubes, find the value of : (i) (12)3+(13)3−(56)3\displaystyle \left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3} (ii) (0.2)3−(0.3)3+(0.1)3\displaystyle (0.2)^{3}-(0.3)^{3}+(0.1)^{3}

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    NCERT’s answer
    (i)
    $\displaystyle -\frac{5}{12}$ (ii) -$\displaystyle 0.018$
    (i) \[\frac12+\frac13-\frac56=0 \] \[\left(\frac12\right)^{3}+\left(\frac13\right)^{3}+\left(-\frac56\right)^{3}=3\left(\frac12\right)\left(\frac13\right)\left(-\frac56\right) \quad \text{(sum zero)} \] \[=-\frac{5}{12} \] (ii) \[0.2-0.3+0.1=0 \] \[(0.2)^{3}+(-0.3)^{3}+(0.1)^{3}=3(0.2)(-0.3)(0.1) \quad \text{(sum zero)} \] \[=-0.018 \] Answer: (i) \(\displaystyle -\dfrac{5}{12}\) (ii) \(\displaystyle -0.018\)

  8. Exercise 38

    Without finding the cubes, factorise (x−2y)3+(2y−3z)3+(3z−x)3(x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}

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    NCERT’s answer
    $\displaystyle 3(x-2 y)(2 y-3 z)(3 z-x)$
    \[(x-2y)+(2y-3z)+(3z-x)=0 \] \[(x-2y)^3+(2y-3z)^3+(3z-x)^3=3(x-2y)(2y-3z)(3z-x) \quad \text{(sum zero)} \]Answer: \(\displaystyle 3(x-2y)(2y-3z)(3z-x)\)

  9. Exercise 39

    Find the value of (i) x3+y3−12xy+64\displaystyle x^{3}+y^{3}-12 x y+64, when x+y=−4\displaystyle x+y=-4 (ii) x3−8y3−36xy−216\displaystyle x^{3}-8 y^{3}-36 x y-216, when x=2y+6\displaystyle x=2 y+6

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    NCERT’s answer
    (i)
    $\displaystyle 0$ (ii) $\displaystyle 0$
    (i) \[x^3+y^3+4^3-3xy(4)=x^3+y^3-12xy+64 \] \[x+y+4=-4+4=0 \] \[\implies x^3+y^3-12xy+64=0 \](ii) \[x^3+(-2y)^3+(-6)^3-3x(-2y)(-6)=x^3-8y^3-36xy-216 \] \[x-2y-6=(2y+6)-2y-6=0 \] \[\implies x^3-8y^3-36xy-216=0 \]Answer: (i) \(\displaystyle 0\) (ii) \(\displaystyle 0\)

  10. Exercise 40

    Give possible expressions for the length and breadth of the rectangle whose area is given by 4a2+4a−3\displaystyle 4 a^{2}+4 a-3.

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    NCERT’s answer
    One possible answer is: Length $\displaystyle =2 a-1$, Breadth $\displaystyle =2 a+3$
    \[4a^2+4a-3=4a^2+6a-2a-3 \] \[=2a(2a+3)-1(2a+3) \] \[=(2a+3)(2a-1) \]Answer: length \(\displaystyle =2a+3\), breadth \(\displaystyle =2a-1\) (or the two interchanged).