Question

Question 38
Without finding the cubes, factorise:
$(x-2y{)}^3+(2y-3z{)}^3+(3z-x{)}^3$

Answer 1

We know that,

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Also, if a + b + c = 0,
then $a^3+b^3+c^3=3abc$ .....condition(1)

Here, we see that
$a+b+c=(x-2y)+(2y-3z)+(3z-x)=0$

$\therefore \text{Using condition (1), we get}\Rightarrow (x-2y{)}^3+(2y-3z{)}^3+(3z-x{)}^3$
$=3(x-2y)(2y-3z)(3z-x)$