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Factorize the expression $(3x+4y)^4-x^4$.
Given:
The given algebraic expression is $(3x+4y)^4-x^4$.
To do:
We have to factorize the expression $(3x+4y)^4-x^4$.
Solution:
Factorizing algebraic expressions:
Factorizing an algebraic expression implies 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.
$(3x+4y)^4-x^4$ can be written as,
$(3x+4y)^4-x^4=[(3x+4y)^2]^2-(x^2)^2$ [Since $(3x+4y)^4=[(3x+4y)^2]^2, x^4=(x^2)^2$]
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,
$(3x+4y)^4-x^4=[(3x+4y)^2]^2-(x^2)^2$
$(3x+4y)^4-x^4=[(3x+4y)^2+x^2][(3x+4y)^2-x^2]$
Now,
Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize $(3x+4y)^2-x^2$.
$(3x+4y)^2-x^2=(3x+4y+x)(3x+4y-x)$
$(3x+4y)^2-x^2=(4x+4y)(2x+4y)$
$(3x+4y)^2-x^2=4(x+y)2(x+2y)$
$(3x+4y)^2-x^2=8(x+y)(x+2y)$.............(I)
Therefore,
$(3x+4y)^4-x^4=[(3x+4y)^2+x^2]8(x+y)(x+2y)$ [Using (I)]
$(3x+4y)^4-x^4=8[(3x+4y)^2+x^2](x+y)(x+2y)$
Hence, the given expression can be factorized as $8[(3x+4y)^2+x^2](x+y)(x+2y)$.
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