Question

Factorize:p^3(q-r)^3+q^3(r-p)^3+r^3(p-q)^3​

Answer 1

Answer:

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

Step-by-step explanation:

Given Equation is p³(q - r)³ + q³(r - p)³ + r³(p - q)³

⇒ [p(q - r)]³ + [q(r - p)]³ + [r(p - q)]³

Let a = p(q - r), b = q(r - p), c = r(p - q)

∴ a + b + c = p(q - r) + q(r - p) + r(p - q)

                 = pq - pr + qr - qp + rp - rq

                 = 0

We know that when a + b + c = 0, then a³ + b³ + c³ = 3abc

⇒ 3(pq - qr)(qr - qp)(rp - rq)

⇒ 3pqr(p - q)(q - r)(r - p)

Hope it helps!

Answer 2

if a + b + c = 0 then factors of a^3 + b^3 + c^3 = 3abc

now p(q -- r) + q(r -- p) + r(p -- q) = 0 hence

{p(q--r)}^3 + {q(r--p)}^3 + {R(p--q)}^3

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

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