Question

Factorize the following: $64a^3-27b^3-144a^{2}b+108a{b}^2$

Answer 1

The expression $64a^3-27b^3-144a^{2}b+108a{b}^2$ can be written as ${\left(4a\right)}^3-{\left(3b\right)}^3-3{\left(4a\right)}^2\left(3b\right)+3\left(4a\right){\left(3b\right)}^2$ ….[1]

We know the Algebraic identity ${\left(a-b\right)}^3=a^3-b^3-3ab\left(a-b\right)$

$=a^3-b^3-3a^{2}b+3a{b}^2$ …[2]

By comparing [1] and [2], we can say $64a^3-27b^3-144a^{2}b+108a{b}^2$$=$${\left(4a\right)}^3-{\left(3b\right)}^3-3{\left(4a\right)}^2\left(3b\right)+3\left(4a\right){\left(3b\right)}^2$

$={\left(4a-3b\right)}^3$

Hence, $\mathbf{64}{\mathit{a}}^{\mathbf{3}}\mathbf{-}\mathbf{27}{\mathit{b}}^{\mathbf{3}}\mathbf{-}\mathbf{144}{\mathit{a}}^{\mathbf{2}}\mathit{b}\mathbf{+}\mathbf{108}\mathit{a}{\mathit{b}}^{\mathbf{2}}\mathbf{=}\left(4a-3b\right)\left(4a-3b\right)\left(4a-3b\right)$