Question

The identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ can be easily verified by expanding the RHS.

Using the above identity (or otherwise) answer the following.

Factorize the expression

$(p+2q-3r{)}^3+(q+2r-3p{)}^3+(r+2p-3q{)}^3$

• A. $(p+q+r)(p^2+q^2+r^2-2(pq+qr+pq\left)\right)$
• B. $(p+q+r)(2p^2+2q^2+2r^2-3(pq+qr+rp\left)\right)$
• C. $3(p+2q-3r)(q+2r-3p)(r+2p-3q)$
• D. None of these

Answer 1

The correct option is C $3(p+2q-3r)(q+2r-3p)(r+2p-3q)$

Let $a=p+2q-3r,b=q+2r-3p,c=r+2p-3q$

$\Rightarrow a+b+c=0$

$\Rightarrow a^3+b^2+c^3-3abc=0$

$\Rightarrow a^3+b^3+c^3=3abc$

$\Rightarrow (p+2q-3r{)}^3+(q+2r-3p{)}^3+(r+2p-3q{)}^3$

$=3(p+2q-3r)(q+2r-3p)(r+2p-3q)$