Question

factorize (p+q)^3-(q+r)^3+(r+p)^3+3(p+q)(q+r)(r+p)​

Answer 1

We know that –

a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca).

here if a + b + c = 0

a3 + b3 + c3 = 3abc.

So, (p-q)3 + (q-r)3 + (r-p)3 = 3(p-q)(q-r)(r-p) {since (p-q) + (q-r) + (r-p) = 0}

Answer 2

Answer:

We know the corollary: if a+b+c=0 then a

3

+b

3

+c

3

=3abc

Using the above corollary taking a=p(q−r), b=q(r−p) and c=r(p−q), we have a+b+c=p(q−r)+q(r−p)+r(p−q)=pq−pr+qr−pq+pr−qr=0 then the equation p

3

(q−r)

3

+q

3

(r−p)

3

+r

3

(p−q)

3

can be factorised as follows:

p

3

(q−r)

3

+q

3

(r−p)

3

+r

3

(p−q)

3

=3[p(q−r)×q(r−p)×r(p−q)]=3pqr(q−r)(r−p)(p−q)

Hence, p

3

(q−r)

3

+q

3

(r−p)

3

+r

3

(p−q)

3

=3pqr(q−r)(r−p)(p−q)