Question

Factorise $125-8x^3-27y^3-90xy$ using the identity $a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]$

Answer 1

The given identity is $a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$

Using the above identity taking $a=5$, $b=-2x$ and $c=-3y$, the equation $125-8x^3-27y^3-90xy$ can be factorised as follows:

$125-8x^3-27y^3-90xy=\frac{1}{2}(5-2x-3y)\left[\right(5-(-2x){)}^2+(-2x-(-3y){)}^2+(-3y-5{)}^2]$

$=\frac{1}{2}(5-2x-3y)\left[\right(5+2x{)}^2+(-2x+3y{)}^2+(-3y-5{)}^2]$

Hence, $125-8x^3-27y^3-90xy=\frac{1}{2}(5-2x-3y)\left[\right(5+2x{)}^2+(-2x+3y{)}^2+(-3y-5{)}^2]$