Question
Factorise:
$1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$
$1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$
✓
Answer
$1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$
$=(1)^3+\Big(\frac{3}{5}\text{a}\Big)^3+3(1)^2\Big(\frac{3}{5}\text{a}\Big)+3(1)\Big(\frac{3}{5}\text{a}\Big)^2$
$=\Big(1+\frac{3}{5}\text{a}\Big)^3$
Hence, factorisation of $1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$ is $=\Big(1+\frac{3}{5}\text{a}\Big)^3$
$=(1)^3+\Big(\frac{3}{5}\text{a}\Big)^3+3(1)^2\Big(\frac{3}{5}\text{a}\Big)+3(1)\Big(\frac{3}{5}\text{a}\Big)^2$
$=\Big(1+\frac{3}{5}\text{a}\Big)^3$
Hence, factorisation of $1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$ is $=\Big(1+\frac{3}{5}\text{a}\Big)^3$
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