Question

If a+b+c=0, then prove that ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$.

Answer 1

Hint: For solving this first we will see the algebraic identity ${{\left( x+y \right)}^{3}}$After that, we will use the given condition and we will prove the desired result.
Complete step-by-step answer:
Given: It is given that value of $a+b+c=0$ and we have to prove that ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$.
Now, before we proceed we should know the following formula:
$\begin{align}
  & {{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right) \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)...................\left( 1 \right) \\
\end{align}$
Now, as we have to prove that ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$ . Now, we will simplify the term ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$ . In the first step, we will apply the formula from the equation (1) with $x=a$ and $y=b$ . Then,
$\begin{align}
  & {{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)+{{c}^{3}} \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b \right)}^{3}}+{{c}^{3}}-3ab\left( a+b \right) \\
\end{align}$
Now, we will use the formula from the equation (1) with $x=a+b$ and $y=c$ . Then,
$\begin{align}
  & {{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b \right)}^{3}}+{{c}^{3}}-3ab\left( a+b \right) \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b+c \right)}^{3}}-3\left( a+b \right)c\left( a+b+c \right)-3ab\left( a+b \right) \\
\end{align}$
Now, write $3ab\left( a+b \right)=3ab\left( a+b+c-c \right)$ in the above equation. Then,
$\begin{align}
  & {{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b+c \right)}^{3}}-3\left( a+b \right)c\left( a+b+c \right)-3ab\left( a+b \right) \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b+c \right)}^{3}}-3\left( a+b \right)c\left( a+b+c \right)-3ab\left( a+b+c-c \right) \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b+c \right)}^{3}}-3\left( a+b \right)c\left( a+b+c \right)-3ab\left( a+b+c \right)+3abc \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=\left( a+b+c \right)\left( {{\left( a+b+c \right)}^{2}}-3\left( a+b \right)c-3ab \right)+3abc \\
\end{align}$
Now, as it is given that $a+b+c=0$. Then,
$\begin{align}
 & \Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=0+3abc \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc \\
\end{align}$
Now, from the above result, we can say that if $a+b+c=0$ then ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$.
Thus, we have proved that the left-hand side term is equal to the right-hand side term.
Note: Here students may go like first will see the algebraic identities like formula for ${{\left( x+y \right)}^{3}}$ , ${{\left( x-y \right)}^{2}}$ and ${{\left( x+y+z \right)}^{2}}$ . After that, we will prove an important algebraic identity ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\dfrac{\left( a+b+c \right)}{2}\left( {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}} \right)$ . And we will use it to prove the desired result. But this is a lengthy solution and it will create confusion.