Question

Factorise: $27a^3+\frac{1}{64b^3}+\frac{27a^2}{4b}+\frac{9a}{16b^2}$.

Answer 1

Consider the given expression

$27a^3+\frac{1}{64b^3}+\frac{27a^2}{4b}+\frac{9a}{16b^2}$ ………..$\left(1\right)$

We know that $(x+y{)}^3=x^3+y^3+3x^{2}y+3x{y}^2+y^3$ …………$\left(2\right)$

Thus,

$27a^3+\frac{1}{64.b^3}+\frac{27.a^2}{4b}+\frac{9a}{16.b^2}$

$=(3a{)}^3+\frac{1}{(4{)}^3,b^3}+3(3a{)}^2.\left(\frac{1}{4b}\right)+3\left(3a\right){\left(\frac{1}{4b}\right)}^2$ ……….$\left(3\right)$

Comparing the above two equations $\left(2\right)$ and $\left(3\right)$, we have

$x=3a;y=\frac{1}{4b}$

Thus equation $\left(1\right)$ becomes.

$27a^3+\frac{1}{64b^3}+\frac{27.a^2}{4b}+\frac{9a}{16.b^2}={[3a+\frac{1}{4b}]}^3$.