Question

Factorise:
$3{(x+y)}^3+\frac{1}{9}{\left(xy\right)}^3$

Answer 1

We know the identity $a^3+b^3=(a+b)(a^2+b^2-ab)$

Using the above identity, the equation can be factorised as follows:

$3(x+y{)}^3+\frac{1}{9}(xy{)}^3=3\left(3\right(x+y{)}^3+\frac{1}{27}\left(xy{)}^3\right)=3\left[\right(x+y{)}^3+{\left(\frac{xy}{3}\right)}^3]$

$=3(x+y+\frac{xy}{3})\left[\right(x+y{)}^2+{\left(\frac{xy}{3}\right)}^2-\left(\right(x+y)\times \frac{xy}{3})]=3(x+y+\frac{xy}{3})(x^2+y^2+2xy-(x+y)\frac{xy}{3}+\frac{x^2{y}^2}{9})$

Hence, $3(x+y{)}^3+\frac{1}{9}(xy{)}^3=3(x+y+\frac{xy}{3})(x^2+y^2+2xy-(x+y)\frac{xy}{3}+\frac{x^2{y}^2}{9})$