Question
Factorise:$125\text{a}^3+\frac{1}{8}$
✓
Answer
$125\text{a}^3+\frac{1}{8}$We know that:
Since $a^2 + b^3 = (a + b)(a^2 - a \times b + b^2)$
Let us rewrite
$125\text{a}^3+\frac{1}{8}$
$=(5\text{a})^3+\Big(\frac{1}{2}\Big)^3$
$=\Big(5\text{a}+\frac{1}{2}\Big)\bigg[(5\text{a})^2-5\text{a}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\bigg]$
$=\Big(5\text{a}+\frac{1}{2}\Big)\Big(25\text{a}^2-\frac{5\text{a}}{2}+\frac{1}{4}\Big)$
Since $a^2 + b^3 = (a + b)(a^2 - a \times b + b^2)$
Let us rewrite
$125\text{a}^3+\frac{1}{8}$
$=(5\text{a})^3+\Big(\frac{1}{2}\Big)^3$
$=\Big(5\text{a}+\frac{1}{2}\Big)\bigg[(5\text{a})^2-5\text{a}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\bigg]$
$=\Big(5\text{a}+\frac{1}{2}\Big)\Big(25\text{a}^2-\frac{5\text{a}}{2}+\frac{1}{4}\Big)$
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