Question

If $a+b+c=0$, then prove that \[\dfrac{{{\left( b+c \right)}^{2}}}{3bc}+\dfrac{{{\left( c+a \right)}^{2}}}{3ca}+\dfrac{{{\left( a+b \right)}^{2}}}{3ab}=1\]

Answer 1

Hint: Put $a+b=-c$
$b+c=-a$
$a+c=-b$, and then solve, and use the identity: “If $a+b+c=0$ then ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$” for further solving.

Complete step-by-step answer:
Given $a+b+c=0.........\left( 1 \right)$

To prove;

\[\dfrac{{{\left( b+c \right)}^{2}}}{3bc}+\dfrac{{{\left( c+a \right)}^{2}}}{3ca}+\dfrac{{{\left( a+b \right)}^{2}}}{3ab}=1.................\left( 2 \right)\]

From equation (1), we can write;

$\begin{align}

  & b+c=-a..............\left( 3 \right) \\

 & a+c=-b...............\left( 4 \right) \\

 & a+b=-c...............\left( 5 \right) \\

\end{align}$

Using equation 3, 4 and 5 in equation (2), we will get;

$\begin{align}

  & \dfrac{{{\left( -a \right)}^{2}}}{3bc}+\dfrac{{{\left( -b \right)}^{2}}}{3ca}+\dfrac{{{\left( -c \right)}^{2}}}{3ab}=1 \\

 & \dfrac{{{a}^{2}}}{3bc}+\dfrac{{{b}^{2}}}{3ca}+\dfrac{{{c}^{2}}}{3ab}=1 \\

\end{align}$

Taking LCM and adding the three terms, equation will be;

$\dfrac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{3abc}=1................\left( 6 \right)$

We know the identity;

“If $a+b+c=0$ then ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$”

If you don’t know this identity, remember this identity is very useful and can be directly used in many questions.

Proof of this identity:

Given: $a+b+c=0$

$\begin{align}

  & \Rightarrow a+b+c=0 \\

 & \Rightarrow a+b=-c \\

\end{align}$

On cubing both sides;

$\Rightarrow {{\left( a+b \right)}^{3}}={{\left( -c \right)}^{3}}$

Identity: ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$

$\Rightarrow {{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)=-{{c}^{3}}$

Using $a+b=-c$, equation will become;

$\begin{align}

  & \Rightarrow {{a}^{3}}+{{b}^{3}}+3ab\left( -c \right)=-{{c}^{3}} \\

 & \Rightarrow {{a}^{3}}+{{b}^{3}}-3abc=-{{c}^{3}} \\

 & \Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc \\

\end{align}$

Using this identity in equation (6);

LHS of equation (6):

$\dfrac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{3abc}=\dfrac{3abc}{3abc}=1$

LHS = RHS = 1

Hence, it is proved that if $a+b+c=0$ then,

\[\dfrac{{{\left( b+c \right)}^{2}}}{3bc}+\dfrac{{{\left( c+a \right)}^{2}}}{3ca}+\dfrac{{{\left( a+b \right)}^{2}}}{3ab}=1\]

Note: Remember the identities used for solving this question.
1- ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$
2- If $a+b+c=0$then ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$