
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
600.9k+ views
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.
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

