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