Question

Factorise: $\frac{1}{64}{a}^3-\frac{1}{16}{a}^{2}b+\frac{1}{12}a{b}^2-\frac{1}{27}{b}^3$

• A. ${(\frac{1}{4}a-\frac{1}{3}b)}^3$
• B. ${(\frac{12}{7}a-\frac{1}{3}b)}^3$
• C. ${(\frac{2}{5}a-\frac{1}{3}b)}^3$
• D. ${(\frac{7}{5}a-\frac{1}{3}b)}^3$

Answer 1

The correct option is A ${(\frac{1}{4}a-\frac{1}{3}b)}^3$

$\frac{1}{64}{a}^3-\frac{1}{16}{a}^{2}b+\frac{1}{12}a{b}^2-\frac{1}{27}{b}^3={\left(\frac{1}{4}a\right)}^3-3\times {\left(\frac{1}{4}a\right)}^2\times \frac{1}{3}b+3\times \frac{1}{4}a\times {\left(\frac{1}{3}b\right)}^2-{\left(\frac{1}{3}b\right)}^3$
$={(\frac{1}{4}a-\frac{1}{3}b)}^3[\because {(a-b)}^3=a^3-3a^{2}b+3a{b}^2-b^3]$