Question

Factorise p^3(q-r)^3+q^3(r-p)^3+r^3(p-q)^3

Answer 1

Step-by-step explanation:

p3(q-r)3+q3(r-p)3+r3(p-q)3

={p(q-r)}3+{q(r-p)}3+{r(p-q)}3

We know, If a+b+c=0 then,

a3+b3+c3 = 3abc

={pq-pr}+{qr-pq}+{pr-qr}

=pq-pr+qr-pq+pr-qr

=0

So, p3(q-r)3+q3(r-p)3+r3(p-q)3=3{p(q-r)q(r-p)r(p-q)}

=3(pq-pr)(qr-pq)(pr-qr)

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Answer 2

Answer:

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

Step-by-step explanation:

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 k = p(q - r), l= q(r - p), m = r(p - q)

∴ k+l+m = 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)