Question

Factorise the following: $\frac{m^3}{64}+\frac{n^3}{27}$

• A. $(\frac{m}{3}+\frac{n}{2})(\frac{m^4}{6}-\frac{n}{12}+\frac{n^3}{9})$
• B. $(\frac{m}{4}+\frac{n}{3})(\frac{m^2}{16}-\frac{mn}{12}+\frac{n^2}{9})$
• C. $(\frac{n}{3}+\frac{m}{3})(\frac{m^2}{16}-\frac{mn}{12}+\frac{n^2}{9})$
• D. $(\frac{m}{5}+\frac{n}{3})(\frac{m^9}{14}-\frac{mn}{12}+\frac{n^2}{9})$

Answer 1

The correct option is B $(\frac{m}{4}+\frac{n}{3})(\frac{m^2}{16}-\frac{mn}{12}+\frac{n^2}{9})$

$\frac{m^3}{64}+\frac{n^3}{27}$

$=(\frac{m}{4}{)}^3+(\frac{n}{3}{)}^3$

Using, $a^3+b^3=(a+b)(a^2-ab+b^2)$

$=[\frac{m}{4}+\frac{n}{3}]\left[\right(\frac{m}{4}{)}^2-\left(\frac{m}{4}\right)\left(\frac{n}{3}\right)+\left(\frac{n}{3}{)}^2\right]$

$=[\frac{m}{4}+\frac{n}{3}][\frac{m^2}{16}-\frac{mn}{12}+\frac{n^2}{9}]$