Question

If a + b + c = 0, prove the identity.
$\left( {{b^2}c + {c^2}a + {a^2}b - 3abc} \right)\left( {b{c^2} + c{a^2} + a{b^2} - 3abc} \right) = {\left( {bc + ca + ab} \right)^2} + 27{a^2}{b^2}{c^2}$

Answer 1

Hint: In this particular question use the standard identities such as ${\left( {ab + bc + ca} \right)^3} = {a^3}{b^3} + {b^3}{c^3} + {c^3}{a^3} + 3abc\left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) + 6{a^2}{b^2}{c^2}$, $\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}} \right) = \left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) + \left( {{a^3} + {b^3} + {c^3}} \right)$ and ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$ so use these identities to reach the solution of the question.

Complete step-by-step answer:
Given data:
a + b + c = 0........................ (1)
Then we have to prove the identity
$\left( {{b^2}c + {c^2}a + {a^2}b - 3abc} \right)\left( {b{c^2} + c{a^2} + a{b^2} - 3abc} \right) = {\left( {bc + ca + ab} \right)^3} + 27{a^2}{b^2}{c^2}$
Consider the LHS of the above equation we have,
$ \Rightarrow \left( {{b^2}c + {c^2}a + {a^2}b - 3abc} \right)\left( {b{c^2} + c{a^2} + a{b^2} - 3abc} \right)$
Now simplify it we have,
$ \Rightarrow \left( {{b^2}c + {c^2}a + {a^2}b} \right)\left( {b{c^2} + c{a^2} + a{b^2}} \right) - 3abc\left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) + 9{a^2}{b^2}{c^2}$..... (2)
Now first simplify $\left( {{b^2}c + {c^2}a + {a^2}b} \right)\left( {b{c^2} + c{a^2} + a{b^2}} \right)$
$ \Rightarrow \left( {{b^2}c + {c^2}a + {a^2}b} \right)\left( {b{c^2} + c{a^2} + a{b^2}} \right) = {b^3}{c^3} + {c^3}{a^3} + {a^3}{b^3} + abc\left( {{a^3} + {b^3} + {c^3}} \right) + 3{a^2}{b^2}{c^2}$........... (3)
Now it is a common known fact that,
 ${\left( {ab + bc + ca} \right)^3} = {a^3}{b^3} + {b^3}{c^3} + {c^3}{a^3} + 3abc\left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) + 6{a^2}{b^2}{c^2}$
$ \Rightarrow {a^3}{b^3} + {b^3}{c^3} + {c^3}{a^3} = {\left( {ab + bc + ca} \right)^3} - 3abc\left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) - 6{a^2}{b^2}{c^2}$
So use this value in equation (3) we have,
\[
   \Rightarrow \left( {{b^2}c + {c^2}a + {a^2}b} \right)\left( {b{c^2} + c{a^2} + a{b^2}} \right){\left( {ab + bc + ca} \right)^3} - 3abc\left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) \\
   - 6{a^2}{b^2}{c^2} + abc\left( {{a^3} + {b^3} + {c^3}} \right) + 3{a^2}{b^2}{c^2} \\
\]
\[
\Rightarrow \left( {{b^2}c + {c^2}a + {a^2}b} \right)\left( {b{c^2} + c{a^2} + a{b^2}} \right) = {\left( {ab + bc + ca} \right)^3} - 3abc\left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) \\
   - 3{a^2}{b^2}{c^2} + abc\left( {{a^3} + {b^3} + {c^3}} \right) \\
 \]
Now substitute this value in equation (2) we have,
$
   \Rightarrow {\left( {ab + bc + ca} \right)^3} - 3abc\left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) - 3{a^2}{b^2}{c^2} + abc\left( {{a^3} + {b^3} + {c^3}} \right) \\
   - 3abc\left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) + 9{a^2}{b^2}{c^2} \\
$
Now simplify it we have,
$ \Rightarrow {\left( {ab + bc + ca} \right)^3} - 6abc\left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) + 6{a^2}{b^2}{c^2} + abc\left( {{a^3} + {b^3} + {c^3}} \right)$....... (4)
Now as we know that,
$\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}} \right) = \left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) + \left( {{a^3} + {b^3} + {c^3}} \right)$
$ \Rightarrow \left( {{b^2}c + {c^2}a + {a^2}b + b{c^2} + c{a^2} + a{b^2}} \right) = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}} \right) - \left( {{a^3} + {b^3} + {c^3}} \right)$
Now substitute this value in equation (4) we have,
$ \Rightarrow {\left( {ab + bc + ca} \right)^3} - 6abc\left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}} \right) - \left( {{a^3} + {b^3} + {c^3}} \right)} \right] + 6{a^2}{b^2}{c^2} + abc\left( {{a^3} + {b^3} + {c^3}} \right)$
Now from equation (1) i.e. a + b + c = 0 we have,
$ \Rightarrow {\left( {ab + bc + ca} \right)^3} + 6abc\left( {{a^3} + {b^3} + {c^3}} \right) + 6{a^2}{b^2}{c^2} + abc\left( {{a^3} + {b^3} + {c^3}} \right)$
Now simplify it we have,
$ \Rightarrow {\left( {ab + bc + ca} \right)^3} + 7abc\left( {{a^3} + {b^3} + {c^3}} \right) + 6{a^2}{b^2}{c^2}$............... (5)
Now it is a common known fact that,
${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$
Now from equation (1) i.e. a + b + c = 0 we have,
$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( 0 \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = 0$
$ \Rightarrow {a^3} + {b^3} + {c^3} = 3abc$
Now substitute this value in equation (5) we have,
$ \Rightarrow {\left( {ab + bc + ca} \right)^3} + 7abc\left( {3abc} \right) + 6{a^2}{b^2}{c^2}$
Now simplify it we have,
$ \Rightarrow {\left( {ab + bc + ca} \right)^3} + 21{a^2}{b^2}{c^2} + 6{a^2}{b^2}{c^2}$
$ \Rightarrow {\left( {ab + bc + ca} \right)^3} + 27{a^2}{b^2}{c^2}$
= RHS
Hence Proved.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic standard identities which is all stated above, these identities are the basis of the solution and always remember that to solve these questions always simplified the given equation using step by step procedure as above simplified, we will get the required result.