Question
Factorise the following:
$8\text{p}^3+\frac{12}{5}\text{p}^2+\frac{6}{25}\text{p}+\frac{1}{125}$
$8\text{p}^3+\frac{12}{5}\text{p}^2+\frac{6}{25}\text{p}+\frac{1}{125}$
✓
Answer
$8\text{p}^3+\frac{12}{5}\text{p}^2+\frac{6}{25}\text{p}+\frac{1}{125}$
$=(2\text{p})^3+3\times(2\text{p})^2\times\frac{1}{5}+3\times(2\text{p})\times\Big(\frac{1}{5}\Big)^2+\Big(\frac{1}{5}\Big)^3$
$=\Big(2\text{p}+\frac{1}{5}\Big)^3$
$\left[\right.$ Using identity, $\left.(a-b)^3=a^3-b^3+3 a(-b)(a-b)\right]$
$=\Big(2\text{p}+\frac{1}{5}\Big)\Big(2\text{p}+\frac{1}{5}\Big)\Big(2\text{p}+\frac{1}{5}\Big)$
$=\Big(2\text{p}+\frac{1}{5}\Big)^3$
$=(2\text{p})^3+3\times(2\text{p})^2\times\frac{1}{5}+3\times(2\text{p})\times\Big(\frac{1}{5}\Big)^2+\Big(\frac{1}{5}\Big)^3$
$=\Big(2\text{p}+\frac{1}{5}\Big)^3$
$\left[\right.$ Using identity, $\left.(a-b)^3=a^3-b^3+3 a(-b)(a-b)\right]$
$=\Big(2\text{p}+\frac{1}{5}\Big)\Big(2\text{p}+\frac{1}{5}\Big)\Big(2\text{p}+\frac{1}{5}\Big)$
$=\Big(2\text{p}+\frac{1}{5}\Big)^3$
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