Question

Question 31
Expand the following:
(i) $(3a–2b{)}^3$

(ii) ${(\frac{1}{x}+\frac{y}{3})}^3$

(iii) ${(4-\frac{1}{3x})}^3$

Answer 1

(i) $(3a–2b{)}^3=(3a{)}^3+(-2b{)}^3+3(3a\left)\right(-2b\left)\right(3a–2b)$
$[Usingtheidentity,(a-b{)}^3=a^3–b^3+3a(-b)(a–b)]$
$=27a^3–8b^3–18ab(3a–2b)$
$=27a^3–8b^3–54a^{2}b+36a{b}^2$
$=27a^3–54a^{2}b+36a{b}^2-8b^3$

(ii) ${(\frac{1}{x}+\frac{y}{3})}^3={\left(\frac{1}{x}\right)}^3+{\left(\frac{y}{3}\right)}^3+3\left(\frac{1}{x}\right)\left(\frac{y}{3}\right)(\frac{1}{x}+\frac{y}{3})$
$[Usingtheidentity,(a+b{)}^3=a^3+b^3+3ab(a+b)]$
$=\frac{1}{x^3}+\frac{y^3}{27}+\frac{y}{x}(\frac{1}{x}+\frac{y}{3})=\frac{1}{x^3}+\frac{y^3}{27}+\frac{y}{x^2}+\frac{y^2}{3x}$

(iii) ${(4-\frac{1}{3x})}^3=(4{)}^3+{(-\frac{1}{3x})}^3+3(4)(-\frac{1}{3x})(4-\frac{1}{3x})$
$[Usingtheidentity,(a-b{)}^3=a^3–b^3+3a(-b)(a–b)]$
$=64-\frac{1}{27x^3}-\frac{4}{x}(4-\frac{1}{3x})$
$=64-\frac{1}{27x^3}-\frac{16}{x}+\frac{4}{3x^2}$