# 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 $8[(3x+4y)^2+x^2](x+y)(x+2y)$.

Updated on: 08-Apr-2023

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