Question

Factorise the following :

$8p^3+\frac{12}{5}{p}^2+\frac{6}{25}p+\frac{1}{125}$

• A. $(p+\frac{1}{5})(2p+\frac{1}{5})(2p+\frac{1}{5})$
• B. $(2p+\frac{1}{5})(2p+\frac{1}{5})(2p-\frac{1}{5})$
• C. $(2p+\frac{1}{5})(2p+\frac{1}{5})(2p+\frac{1}{5})$
• D. $(p+\frac{1}{5})(2p+\frac{1}{5})(2p-\frac{1}{5})$

Answer 1

The correct option is C $(2p+\frac{1}{5})(2p+\frac{1}{5})(2p+\frac{1}{5})$

$8p^3+\frac{12}{5}{p}^2+\frac{6}{25}p+\frac{1}{125}$

$=8p^3+\frac{1}{125}+\frac{6}{5}p(2p+\frac{1}{5})$

$=(2p{)}^3+(\frac{1}{5}{)}^3+3\left(2p\right)\left(\frac{1}{5}\right)(2p+\frac{1}{5})$$\left[\right(a+b{)}^3=a^3+b^3+3ab(a+b)]$

$=(2p+\frac{1}{5}{)}^3$

$=(2p+\frac{1}{5})(2p+\frac{1}{5})(2p+\frac{1}{5})$