# Factorize the expression $a^4b^4-81c^4$.

Given:

The given expression is $a^4b^4-81c^4$.

To do:

We have to factorize the expression $a^4b^4-81c^4$.

Solution:

Factorizing algebraic expressions:

Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution.

An algebraic expression is factored completely when it is written as a product of prime factors.

$a^4b^4-81c^4$ can be written as,

$a^4b^4-81c^4=(a^2b^2)^2-(9c^2)^2$             [Since $a^4b^4=(a^2b^2)^2, 81c^4=(9c^2)^2$]

Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression.

Therefore,

$a^4b^4-81c^4=(a^2b^2)^2-(9c^2)^2$

$a^4b^4-81c^4=(a^2b^2+9c^2)(a^2b^2-9c^2)$

Now,

$a^2b^2-9c^2$ can be written as,

$a^2b^2-9c^2=(ab)^2-(3c)^2$                         (Since $a^2b^2=(ab)^2, 9c^2=(3c)^2$)

Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize $(ab)^2-(3c)^2$.

$(ab)^2-(3c)^2=(ab+3c)(ab-3c)$.............(I)

Therefore,

$a^4b^4-81c^4=(a^2b^2+9c^2)(ab+3c)(ab-3c)$                 [Using (I)]

Hence, the given expression can be factorized as $(a^2b^2+9c^2)(ab+3c)(ab-3c)$.

Updated on: 09-Apr-2023

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