Question

Factorise: $\frac{a^3}{8}+\frac{b^3}{27}+\frac{a^{2}b}{4}+\frac{a{b}^2}{6}$

• A. $(\frac{a}{2}+\frac{b}{3})(\frac{a}{2}+\frac{b}{3})(\frac{a}{2}+\frac{b}{3})$
• B. $(\frac{a}{2}+\frac{b}{5})(\frac{a}{2}+\frac{b}{3})(\frac{a}{2}+\frac{b}{11})$
• C. $(\frac{a}{2}+\frac{b}{7})(\frac{a}{2}+\frac{b}{3})(\frac{a}{2}+\frac{b}{6})$
• D. $(\frac{a}{2}+\frac{b}{9})(\frac{a}{2}+\frac{b}{3})(\frac{a}{2}+\frac{b}{3})$

Answer 1

The correct option is A $(\frac{a}{2}+\frac{b}{3})(\frac{a}{2}+\frac{b}{3})(\frac{a}{2}+\frac{b}{3})$

$\frac{a^3}{8}+\frac{b^3}{27}+\frac{a^{2}b}{4}+\frac{a{b}^2}{6}=\frac{a^3}{2^3}+\frac{b^3}{3^3}+3\times \frac{a^2}{2^2}\times \frac{b}{3}+3\times \frac{a}{2}\times \frac{b^2}{3^2}=\frac{a^3}{2^3}+\frac{b^3}{3^3}+3\times \frac{a}{2}\times \frac{b}{3}(\frac{a}{2}+\frac{b}{3})$
Using,$(a+b{)}^3=a^3+b^3+3ab(a+b)$
$={(\frac{a}{2}+\frac{b}{3})}^3$
$=(\frac{a}{2}+\frac{b}{3})(\frac{a}{2}+\frac{b}{3})(\frac{a}{2}+\frac{b}{3})$