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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 $(a^2b^2+9c^2)(ab+3c)(ab-3c)$.
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