RS AGGARWAL CLASS 9 Chapter 3 FACTORISATION OF POLYNOMIAL EXERCISE 3E

  EXERCISE 3E

PAGE NO-123

Question 1:

Expand
(i) (3x + 2)³
(ii) ${\left(3a+\frac{1}{4b}\right)}^3$
(iii) ${\left(1+\frac{2}{3}a\right)}^3$

Answer 1:

$$\begin{gathered} \left(\mathrm{i}\right) {\left(3x+2\right)}^3={\left(3x\right)}^3+3\times {\left(3x\right)}^{2}x2+3\times 3x\times {\left(2\right)}^2+{\left(2\right)}^3 \\ =27x^3+54x^2+36x+8 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right) {\left(3a+\frac{1}{4b}\right)}^3={\left(3a\right)}^3+{\left(\frac{1}{4b}\right)}^3+3{\left(3a\right)}^2\left(\frac{1}{4b}\right)+3\left(3a\right){\left(\frac{1}{4b}\right)}^2 \\ =27a^3+\frac{1}{64b^3}+\frac{27a^2}{4b}+\frac{9a}{16b^2} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) {\left(1+\frac{2}{3}a\right)}^3={\left(\frac{2}{3}a\right)}^3+3\times {\left(\frac{2}{3}a\right)}^2\times 1+3a\frac{2}{3}a\times {\left(1\right)}^2+{\left(1\right)}^3 \\ =\frac{8}{27}{a}^3+\frac{4}{3}{a}^2+2a+1 \end{gathered}$$

Question 2:

Expand
(i) (5a – 3b)³
(ii) ${\left(3x-\frac{5}{x}\right)}^3$
(iii) ${\left(\frac{4}{5}a-2\right)}^3$

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) {\left(5a-3b\right)}^3={\left(5a\right)}^3-{\left(3b\right)}^3-3{\left(5a\right)}^2\left(3b\right)+3\left(5a\right){\left(3b\right)}^2 \\ =125a^3-27b^3-225a^{2}b+135a{b}^2 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) {\left(3x-\frac{5}{x}\right)}^3={\left(3x\right)}^3-{\left(\frac{5}{x}\right)}^3-3{\left(3x\right)}^2\left(\frac{5}{x}\right)+3\left(3x\right){\left(\frac{5}{x}\right)}^2 \\ =27x^3-\frac{125}{x^3}-135x+\frac{225}{x} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right) {\left(\frac{4}{5}a-2\right)}^3={\left(\frac{4}{5}a\right)}^3-{\left(2\right)}^3-3{\left(\frac{4}{5}a\right)}^2\left(2\right)+3\left(\frac{4}{5}a\right){\left(2\right)}^2 \\ =\frac{64}{125}{a}^3-8-\frac{96}{25}{a}^2+\frac{48}{5}a \end{gathered}$$

Question 3:

Factorise
$8a^3+27b^3+36a^{2}b+54a{b}^2$

Answer 3:

$$\begin{gathered} 8a^3+27b^3+36a^{2}b+54a{b}^2={\left(2a\right)}^3+{\left(3b\right)}^3+3{\left(2a\right)}^2\left(3b\right)+3\left(2a\right){\left(3b\right)}^2 \\ ={\left(2a+3b\right)}^3 \end{gathered}$$

Hence, factorisation of $8a^3+27b^3+36a^{2}b+54a{b}^2$ is ${\left(2a+3b\right)}^3$.

Question 4:

Factorise
$64a^3-27b^3-144a^{2}b+108a{b}^2$

Answer 4:

$$\begin{gathered} 64a^3-27b^3-144a^{2}b+108a{b}^2={\left(4a\right)}^3-{\left(3b\right)}^3-3{\left(4a\right)}^2\left(3b\right)+3\left(4a\right){\left(3b\right)}^2 \\ ={\left(4a-3b\right)}^3 \end{gathered}$$

Hence, factorisation of $64a^3-27b^3-144a^{2}b+108a{b}^2$ is ${\left(4a-3b\right)}^3$.

Question 5:

Factorise
$1+\frac{27}{125}{a}^3+\frac{9a}{5}+\frac{27a^2}{24}$

Answer 5:

$$\begin{gathered} 1+\frac{27}{125}{a}^3+\frac{9a}{5}+\frac{27a^2}{25}={\left(1\right)}^3+{\left(\frac{3}{5}a\right)}^3+3{\left(1\right)}^2\left(\frac{3}{5}a\right)+3\left(1\right){\left(\frac{3}{5}a\right)}^2 \\ ={\left(1+\frac{3}{5}a\right)}^3 \end{gathered}$$

Hence, factorisation of $1+\frac{27}{125}{a}^3+\frac{9a}{5}+\frac{27a^2}{25}$ is ${\left(1+\frac{3}{5}a\right)}^3$.

Question 6:

Factorise
$125x^3-27y^3-225x^{2}y+135x{y}^2$

Answer 6:

$$\begin{gathered} 125x^3-27y^3-225x^{2}y+135x{y}^2={\left(5x\right)}^3-{\left(3y\right)}^3-3{\left(5x\right)}^2\left(3y\right)+3\left(5x\right){\left(3y\right)}^2 \\ ={\left(5x-3y\right)}^3 \end{gathered}$$

Hence, factorisation of $125x^3-27y^3-225x^{2}y+135x{y}^2$ is ${\left(5x-3y\right)}^3$.

Question 7:

Factorise
$a^3{x}^3-3a^{2}b{x}^2+3a{b}^{2}x-b^3$

Answer 7:

$$\begin{gathered} a^3{x}^3-3a^{2}b{x}^2+3a{b}^{2}x-b^3={\left(ax\right)}^3-{\left(b\right)}^3-3{\left(ax\right)}^2\left(b\right)+3\left(ax\right){\left(b\right)}^2 \\ ={\left(ax-b\right)}^3 \end{gathered}$$

Hence, factorisation of $a^3{x}^3-3a^{2}b{x}^2+3a{b}^{2}x-b^3$ is ${\left(ax-b\right)}^3$.

Question 8:

Factorise
$\frac{64}{125}{a}^3-\frac{96}{25}{a}^2+\frac{48}{5}a-8$

Answer 8:

$$\begin{gathered} \frac{64}{125}{a}^3-\frac{96}{25}{a}^2+\frac{48}{5}a-8={\left(\frac{4}{5}a\right)}^3-{\left(2\right)}^3-3{\left(\frac{4}{5}a\right)}^2\left(2\right)+3\left(\frac{4}{5}a\right){\left(2\right)}^2 \\ ={\left(\frac{4}{5}a-2\right)}^3 \end{gathered}$$

Hence, factorisation of $\frac{64}{125}{a}^3-\frac{96}{25}{a}^2+\frac{48}{5}a-8$ is ${\left(\frac{4}{5}a-2\right)}^3$.

Question 9:

Factorise
a³ – 12a(a – 4) – 64

Answer 9:

$$\begin{gathered} a^3-12a\left(a-4\right)-64=a^3-12a^2+48a-64 \\ ={\left(a\right)}^3-{\left(4\right)}^3-3{\left(a\right)}^2\left(4\right)+3\left(a\right){\left(4\right)}^2 \\ ={\left(a-4\right)}^3 \end{gathered}$$

Hence, factorisation of a³ – 12a(a – 4) – 64 is ${\left(a-4\right)}^3$.

Question 10:

Evaluate
(i) (103)³
(ii) (99)³

Answer 10:

$$\begin{gathered} \left(\mathrm{i}\right) {\left(103\right)}^3={\left(100+3\right)}^3 \\ ={\left(100\right)}^3+{\left(3\right)}^3+3{\left(100\right)}^2\left(3\right)+3\left(100\right){\left(3\right)}^2 \\ =1000000+27+90000+2700 \\ =1092727 \\ \left(\mathrm{ii}\right) {\left(99\right)}^3={\left(100-1\right)}^3 \\ ={\left(100\right)}^3-{\left(1\right)}^3-3{\left(100\right)}^2\left(1\right)+3\left(100\right){\left(1\right)}^2 \\ =1000000-1-30000+300 \\ =1000300-30001 \\ =970299 \end{gathered}$$