Question

If $a+b+c=0$, prove that identities $(b^{2}c+c^{2}a+a^{2}b-3abc)(b{c}^2+c{a}^2+a{b}^2-3abc)=(bc+ca+ab{)}^3+27a^2{b}^2{c}^2$.

Answer 1

$\Rightarrow (b^{2}c+c^{2}a+a^{2}b-3abc)(b{c}^2+c{a}^2+a{b}^2-3abc)$

$=(b^{2}c+c^{2}a+a^{2}b)(b{c}^2+c{a}^2+a{b}^2)-3abc\sum a^{2}b+9a^2{b}^2{c}^2⟶1$

Now, $(b^{2}c+c^{2}a+a^{2}b)(b{c}^2+c{a}^2+a{b}^2)=b^3{c}^3+c^3{a}^3+a^3{b}^3+abc(a^3+b^3+c^3)+3a^2{b}^2{c}^2$

$=\{{(bc+ca+ab)}^3-3abc\sum a^{2}b-6a^2{b}^2{c}^2\}+abc(a^3+b^3+c^3)+3a^2{b}^2{c}^2$

Hence from $1$ the given expression,

$={(bc+ca+ab)}^3-6abc\sum a^{2}b+6a^2{b}^2{c}^2+abc(a^3+b^3+c^3)$

But $\sum a^{2}b=(a+b+c)(a^2+b^2+c^2)-3abc$

$=-3abc\because a+b+c=0$

Also, $a^3+b^3+c^3=3abc$

Hence the expression,

$={(bc+ca+ab)}^3+18a^2{b}^2{c}^2+6a^2{b}^2{c}^2+3a^2{b}^2{c}^2$

$={(bc+ca+ab)}^3+27a^2{b}^2{c}^2$