Question

Factorize: $(x-2y{)}^3+(2y-3z{)}^3+(3z-x{)}^3$.

Answer 1

Here to factorise $(x-24{)}^3+(2y-3z{)}^3+(3z-x{)}^3$, we need to assume an another equation

$\Rightarrow$ Therefore

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab+bc+ca).....\left(1\right)$

And in the above equation is

$a+b+c=0$, then

$a^3+b^3+c^3=3abc....\left(2\right)$

$\to$ So, for the above question if we assume

$a=(x-2y)\to \left(3\right)$

$b=(2y-3z)\to \left(4\right)$

$c=(3z-x)\to \left(5\right)$

Then,

$(x-2y)+(2y-3z)+(3z-x)=0\to \left(6\right)$

$[\because a+b+c=0]$

Therefore,

$(x-2y{)}^3+(2y-3z{)}^3+(3z-x{)}^3=3(x-2y\left)\right(2y-3z\left)\right(3z-x)$