Question

Factorise the expression $\sum{a\left( {{b}^{2}}-{{c}^{2}} \right)}$.
A. $-\left( a-b \right)\left( b-c \right)\left( c-a \right)$
B. $\left( a-b \right)\left( b-c \right)\left( c-a \right)$
C. $\left( a-b \right)\left( b-c \right)\left( c+a \right)$
D. None

Answer 1

Hint: For this type of questions, you should know about factorisation by expanding the brackets and then taking the commons again from the solving step by step to the question. The brackets which will be expanded in it will be expanded in the following ways: for an expression of the form $x\left( y+z \right)$, the expanded version of this will be $xy+xz$.

Complete step-by-step solution:
Now according to the question, we have to factorise $\sum{a\left( {{b}^{2}}-{{c}^{2}} \right)}$, which can be written as,
$a\left( {{b}^{2}}-{{c}^{2}} \right)+b\left( {{c}^{2}}-{{a}^{2}} \right)+c\left( {{a}^{2}}-{{b}^{2}} \right)$
So, now we will expand the brackets and solve this question. By expanding the brackets, we will get,
$a{{b}^{2}}-a{{c}^{2}}+b{{c}^{2}}-b{{a}^{2}}+c{{a}^{2}}-c{{b}^{2}}$
By taking the similar values with one another, we will get,
$\begin{align}
  & a{{b}^{2}}-{{a}^{2}}b+{{a}^{2}}c-c{{b}^{2}}+b{{c}^{2}}-a{{c}^{2}} \\
 & \Rightarrow ab\left( b-a \right)+c\left( {{a}^{2}}-{{b}^{2}} \right)+{{c}^{2}}\left( b-a \right) \\
 & \because {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \\
\end{align}$
We can write the above equation as follows,
$ab\left( b-a \right)-c\left( a+b \right)\left( b-a \right)+{{c}^{2}}\left( b-a \right)$
By taking $\left( b-a \right)$ common form all, we will get,
$\begin{align}
  & \left( b-a \right)\left[ ab-c\left( a+b \right)+{{c}^{2}} \right] \\
 & \Rightarrow \left( b-a \right)\left[ ab-ca-cb+{{c}^{2}} \right] \\
\end{align}$
Now taking the common terms in the bracket, we will get,
$\begin{align}
  & \left( b-a \right)\left[ a\left( b-c \right)+c\left( c-b \right) \right] \\
 & \Rightarrow \left( b-a \right)\left[ \left( b-c \right)a-c\left( b-c \right) \right] \\
\end{align}$
By solving this bracket, we will get,
$\begin{align}
  & \left( b-a \right)\left( b-c \right)\left( a-c \right) \\
 & \Rightarrow \left( a-b \right)\left( b-c \right)\left( c-a \right) \\
\end{align}$
So, the answer for factoring of $\sum{a\left( {{b}^{2}}-{{c}^{2}} \right)}$ is $\left( a-b \right)\left( b-c \right)\left( c-a \right)$.
Hence the correct option is option B.

Note: In factorising we also do the reverse of the expanding of the brackets, which is the very important most step in solving the quadratic equations. In this step we have to see that if any terms have any one common factor, then we will take them out and then make new equations which will very much help in solving the equation.