Question & Answer

Factorize the following equation : ${a^4} - {b^4}$
A. $\left( {{a^2} + {b^2}} \right)\left( {a - b} \right)\left( {a - b} \right)$
B. $\left( {{a^2} + {b^2}} \right)\left( {a - b} \right)\left( {4a - b} \right)$
C. $\left( {{a^2} + {b^2}} \right)\left( {a + b} \right)\left( {4a - b} \right)$
D. $\left( {{a^2} + {b^2}} \right)\left( {a + b} \right)\left( {a - b} \right)$

ANSWER Verified Verified
Hint: In order to factorize given term, we will try to reduce it with help of formula $\left( {{x^2} - {y^2}} \right) = \left( {x + y} \right)\left( {x - y} \right)$. Again reduce the equation by using the same formula.

Complete step-by-step answer:
Given term is ${a^4} - {b^4}$
This term can be written in the form of ${a^2}$ by breaking so
We know that ${\left( {{x^p}} \right)^q} = {x^{pq}}$
   \Rightarrow {a^{2 \times 2}} - {b^{2 \times 2}} \\
   \Rightarrow {\left( {{a^2}} \right)^2} - {\left( {{b^2}} \right)^2}{\text{ }}\left[ {{\text{by using the above formula }}} \right] \\
Let ${a^2} = p$ and ${b^2} = q,$ so we have
   = {p^2} - {q^2} \\
  \because {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) \\
Substitute the value of x as p and y as q, we have
$ = \left( {p - q} \right)\left( {p + q} \right)$
Now, put the value of $p$ and $q$ as ${a^2}{\text{ and }}{b^2}$ we get
$ = \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right)$
Again, we will apply same formula as
\[{x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\] we obtain
\[ = \left( {a - b} \right)\left( {a + b} \right)\left( {{a^2} + {b^2}} \right)\]
Hence, the required answer is $\left( {a - b} \right)\left( {a + b} \right)\left( {{a^2} + {b^2}} \right)$ and ”D” is the correct option.

Note: In order to solve these types of questions, remember the basic formula of all algebraic equations and try to solve the given expression step by step. The point to remember while factoring the expression is take out the common terms if any, see if the expression fits any of the identities, plus any more you know and keep solving the expression for your answer.