Question

State True or False.

$27p^3-\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4}p$$=(3p-\frac{1}{6})(3p-\frac{1}{6})(3p-\frac{1}{6})$

• A. True
• B. False

Answer 1

The correct option is A True

We have, $27p^3-\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4}p$
Using, $(a+b{)}^3=a^3+b^3+3ab(a+b)$
$=(3p{)}^3+(-\frac{1}{6}{)}^3+3\left(3p\right)(-\frac{1}{6})(3p-\frac{1}{6})$
$=(3p-\frac{1}{6}{)}^3$
$=(3p-\frac{1}{6})(3p-\frac{1}{6})(3p-\frac{1}{6})$