Question

If we have an expression $a+b+c=0$, prove the identity ${{a}^{6}}+{{b}^{6}}+{{c}^{6}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-2{{\left( ab+bc+ac \right)}^{3}}$.

Answer 1

Hint: For answering this question we need to prove the equation ${{a}^{6}}+{{b}^{6}}+{{c}^{6}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-2{{\left( ab+bc+ac \right)}^{3}}$using the value $a+b+c=0$. Here from the given value we can obtain$a+b=-c\And b+c=-a\And c+a=-b$. We also know that ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$ as $a+b+c=0$ we can say that
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$ by squaring and simplifying this using ${{\left( a+b+c \right)}^{3}}={{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3\left( a+b \right)\left( b+c \right)\left( c+a \right)$ and ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$.

Complete step-by-step solution
Given in the question that $a+b+c=0$
We know that ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$
So we can say that ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$
By squaring this equation we get ${{\left( {{a}^{3}}+{{b}^{3}}+{{c}^{3}} \right)}^{2}}={{\left( 3abc \right)}^{2}}$
Since we know that${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$we will use this identity while expanding${{\left( {{a}^{3}}+{{b}^{3}}+{{c}^{3}} \right)}^{2}}={{\left( 3abc \right)}^{2}}$.
By expanding this we will get
${{a}^{6}}+{{b}^{6}}+{{c}^{6}}+2{{a}^{3}}{{b}^{3}}+2{{b}^{3}}{{c}^{3}}+2{{c}^{3}}{{a}^{3}}=9{{a}^{2}}{{b}^{2}}{{c}^{2}}$
As we need to prove ${{a}^{6}}+{{b}^{6}}+{{c}^{6}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-2{{\left( ab+bc+ac \right)}^{3}}$
We will transfer some parts from L.H.S to R.H.S for simplifications
${{a}^{6}}+{{b}^{6}}+{{c}^{6}}=9{{a}^{2}}{{b}^{2}}{{c}^{2}}-2\left( {{a}^{3}}{{b}^{3}}+{{b}^{3}}{{c}^{3}}+{{c}^{3}}{{a}^{3}} \right)$
As we need to prove ${{a}^{6}}+{{b}^{6}}+{{c}^{6}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-2{{\left( ab+bc+ac \right)}^{3}}$ we will compare this with the equation we have and do the further required modifications.
$\Rightarrow {{a}^{6}}+{{b}^{6}}+{{c}^{6}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}+6{{a}^{2}}{{b}^{2}}{{c}^{2}}-2\left( {{a}^{3}}{{b}^{3}}+{{b}^{3}}{{c}^{3}}+{{c}^{3}}{{a}^{3}} \right)$
Since it is given in the question that $a+b+c=0$
We can expand $6{{a}^{2}}{{b}^{2}}{{c}^{2}}$as $-6b\left( a+c \right)c\left( b+a \right)a\left( c+b \right)$ because $a+b=-c\And b+c=-a\And c+a=-b$.
After expanding using $6{{a}^{2}}{{b}^{2}}{{c}^{2}}$=$-6b\left( a+c \right)c\left( b+a \right)a\left( c+b \right)$ we will have
$\begin{align}
  & \Rightarrow {{a}^{6}}+{{b}^{6}}+{{c}^{6}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-6b\left( a+c \right)c\left( b+a \right)a\left( c+b \right)-2\left( {{a}^{3}}{{b}^{3}}+{{b}^{3}}{{c}^{3}}+{{c}^{3}}{{a}^{3}} \right) \\
 & \Rightarrow {{a}^{6}}+{{b}^{6}}+{{c}^{6}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-2\left( \left( {{a}^{3}}{{b}^{3}}+{{b}^{3}}{{c}^{3}}+{{c}^{3}}{{a}^{3}} \right)+3\left( ab+cb \right)\left( bc+ac \right)\left( ac+ab \right) \right) \\
\end{align}$
In the above equation we had taken 2 as common.
As we know that ${{\left( a+b+c \right)}^{3}}={{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3\left( a+b \right)\left( b+c \right)\left( c+a \right)$
We can say that ${{\left( ab+bc+ca \right)}^{3}}=\left( \left( {{a}^{3}}{{b}^{3}}+{{b}^{3}}{{c}^{3}}+{{c}^{3}}{{a}^{3}} \right)+3\left( ab+cb \right)\left( bc+ac \right)\left( ac+ab \right) \right)$
So we can simply write the above equation as
$\Rightarrow {{a}^{6}}+{{b}^{6}}+{{c}^{6}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-2\left( \left( {{a}^{3}}{{b}^{3}}+{{b}^{3}}{{c}^{3}}+{{c}^{3}}{{a}^{3}} \right)+3\left( ab+cb \right)\left( bc+ac \right)\left( ac+ab \right) \right)$
$\Rightarrow {{a}^{6}}+{{b}^{6}}+{{c}^{6}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-2{{\left( ab+bc+ca \right)}^{3}}$
Hence, the required identity ${{a}^{6}}+{{b}^{6}}+{{c}^{6}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-2{{\left( ab+bc+ca \right)}^{3}}$ is proved when$a+b+c=0$.

Note: While performing proofs the equation required to prove should be in our mind otherwise we will end up having a messy loop, which will waste over time. The simplifications should be done in a method to obtain the required equations.