Question

Find the factors of ${{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right).$

Answer 1

Hint: For factorising the above expression, split the middle term as sum or difference of other two terms and use the identity $''{{x}^{2}}-{{y}^{2}}=\left( x-y \right)\left( x+y \right)''$ for further factorization. Split ${{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)=-{{b}^{4}}\left[ \left( {{b}^{2}}-{{c}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right) \right]$ and further factorise the obtained algebraic expression.

Complete step-by-step answer:
The given equation \[{{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right)\].
Let us split the middle term ${{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)={{b}^{4}}\left[ -\left( \left( {{b}^{2}}-{{c}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right) \right) \right]$
$\begin{align}
  & \left[ -\left( \left( {{b}^{2}}-{{c}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right) \right) \right]=\left[ -\left({{{b}^{2}}}-{{c}^{2}}+{{a}^{2}}-{{{b}^{2}}} \right) \right] \\
 & =\left[ -\left( {{a}^{2}}-{{c}^{2}} \right) \right] \\
 & \Rightarrow \left[ -\left( \left( {{b}^{2}}-{{c}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right) \right) \right]={{c}^{2}}-{{a}^{2}} \\
\end{align}$
Splitting the middle term and writing ${{c}^{2}}-{{a}^{2}}=\left[ -\left( \left( {{b}^{2}}-{{c}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right) \right) \right]$, we will get, $\begin{align}
  & {{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right)={{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left[ -\left( {{b}^{2}}-{{c}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right) \right]+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right) \\
 & ={{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)-{{b}^{4}}\left( \left( {{b}^{2}}-{{c}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right) \right) \\
 & ={{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)-{{b}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)-{{b}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right) \\
\end{align}$Taking $\left( {{b}^{2}}-{{c}^{2}} \right)$ common from first two terms and $\left( {{a}^{2}}-{{b}^{2}} \right)$ common from last two terms, we will get;
$\begin{align}
  & {{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right)=\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{a}^{4}}-{{b}^{4}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{c}^{4}}-{{b}^{4}} \right) \\
 & =\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{a}^{4}}-{{b}^{4}} \right)-\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{b}^{4}}-{{c}^{4}} \right) \\
\end{align}$
We know that${{x}^{2}}-{{y}^{2}}=\left( x-y \right)\left( x+y \right)$.
$\Rightarrow \left( {{b}^{4}}-{{c}^{4}} \right)={{\left( {{b}^{2}} \right)}^{2}}-{{\left( {{c}^{2}} \right)}^{2}}$
$=\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}} \right)$
Replacing $\left( {{b}^{4}}-{{c}^{4}} \right)=\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}} \right)$ in the last term of RHS, we will get;
${{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right)=\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{a}^{4}}-{{b}^{4}} \right)-\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}} \right)$
Similarly, ${{a}^{4}}-{{b}^{4}}=\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)$
Replacing ${{a}^{4}}-{{b}^{4}}=\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)$ in RHS, we will get;
${{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right)=\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)-\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}} \right)$ Taking $\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{b}^{2}}-{{c}^{2}} \right)$ common from both the terms in RHS, we will get;
$\begin{align}
  & {{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right)=\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{a}^{2}}-{{b}^{2}} \right)\left[ \left( {{a}^{2}}+{{b}^{2}} \right)-\left( {{b}^{2}}+{{c}^{2}} \right) \right] \\
 & \Rightarrow {{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right)=\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{a}^{2}}-{{b}^{2}} \right)\left[ {{a}^{2}}+{{{b}^{2}}}-{{{b}^{2}}}-{{c}^{2}} \right] \\
 & =\left( {{b}^{2}}-{{c}^{2}} \right)\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}-{{c}^{2}} \right) \\
\end{align}$
As we know $\left( {{x}^{2}}-{{y}^{2}} \right)=\left( x-y \right)\left( x+y \right)$,
$\begin{align}
  & \left( {{b}^{2}}-{{c}^{2}} \right)=\left( b-c \right)\left( b+c \right)\ and \\
 & \left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)\ and \\
 & \left( {{a}^{2}}-{{c}^{2}} \right)=\left( a-c \right)\left( a+c \right) \\
\end{align}$
Using these in above equation, we will get;
 ${{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right)=\left( b-c \right)\left( b+c \right)\left( a-b \right)\left( a+b \right)\left( a-c \right)\left( a+c \right)$
Hence, ${{a}^{4}}\left( {{b}^{2}}-{{c}^{2}} \right)+{{b}^{4}}\left( {{c}^{2}}-{{a}^{2}} \right)+{{c}^{4}}\left( {{a}^{2}}-{{b}^{2}} \right)$ can be factorised into $\left( b-c \right)\left( b+c \right)\left( a-b \right)\left( a+b \right)\left( a-c \right)\left( a+c \right)$.

Note:Students can do mistake by initially expanding the expression by multiplying ${{a}^{4}},{{b}^{4}}\ and\ {{c}^{4}}$ with their corresponding terms and then factorising. After expanding, you will get a biquadratic expression whose factorisation will be tough. So, middle term splitting is the best method that can be used here.