Question

Fatorise: $\frac{1}{8}{a}^3-\frac{3}{16}{a}^2+\frac{3}{32}a-\frac{1}{64}$

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

Answer 1

The correct option is D $(\frac{a}{2}-\frac{1}{4})(\frac{a}{2}-\frac{1}{4})(\frac{a}{2}-\frac{1}{4})$

$\frac{1}{8}{a}^3-\frac{3}{16}{a}^2+\frac{3}{32}a-\frac{1}{64}$
$=\frac{1}{2^3}{a}^3-\frac{1}{4^3}+3\times \frac{a^2}{2^2}\times \frac{1}{4}+3\times \frac{a}{2}\times \frac{1}{4^2}$
$=\frac{1}{2^3}{a}^3-\frac{1}{4^3}+3\times \frac{a}{2}\times \frac{1}{4}(\frac{a}{2}+\frac{1}{4})$
Using,$(a-b{)}^3=a^3-b^3-3ab(a-b)$
$={(\frac{a}{2}-\frac{1}{4})}^3$
$=(\frac{a}{2}-\frac{1}{4})(\frac{a}{2}-\frac{1}{4})(\frac{a}{2}-\frac{1}{4})$