# Factorize the expression $a^4-16(b-c)^4$.

Given:

The given expression is $a^4-16(b-c)^4$.

To do:

We have to factorize the expression $a^4-16(b-c)^4$.

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.

$a^4-16(b-c)^4$ can be written as,

$a^4-16(b-c)^4=(a^2)^2-[4(b-c)^2]^2$             [Since $a^4=(a^2)^2, 16(b-c)^4=(4(b-c)^2)^2$]

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,

$a^4-16(b-c)^4=(a^2)^2-[4(b-c)^2]^2$

$a^4-16(b-c)^4=[a^2+4(b-c)^2][a^2-4(b-c)^2]$

Now,

$a^2-4(b-c)^2$ can be written as,

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

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

$(a)^2-[2(b-c)]^2=[a+2(b-c)][a-2(b-c)]$

$(a)^2-[2(b-c)]^2=(a+2b-2c)(a-2b+2c)$.............(I)

Therefore,

$a^4-16(b-c)^4=[a^2+4(b-c)^2](a+2b-2c)(a-2b+2c)$                 [Using (I)]

Hence, the given expression can be factorized as $[a^2+4(b-c)^2](a+2b-2c)(a-2b+2c)$.

Updated on: 09-Apr-2023

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