Question

pls tell i will mark u as brainlist​

Answer 1

Required Knowledge

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

This is the required identity because the sum of three terms without exponent is zero.

What happens if $a+b+c=0$? The LHS becomes zero since it is a product with zero. Then we are left with $a^3+b^3+c^3-3abc=0$.

$\implies \boxed{a^3+b^3+c^3=3abc}$

Solution

First, we are going to distribute the terms. We see that the sum of $(pq-rp)+(qr-pq)+(rp-qr)$ is exactly zero. Then, we see that $a^3+b^3+c^3=3abc$.

Given: $p^3(q-r)^3+q^3(r-p)^3+r^3(p-q)^3$

$=(pq-rp)^3+(qr-pq)^3+(rp-qr)^3$

$=3(pq-rp)(qr-pq)(rp-qr)$

$=3pqr(p-q)(q-r)(r-p)$

Hence factorized.