Factorize the expression $16(a-b)^3-24(a-b)^2$.

Given:

The given expression is $16(a-b)^3-24(a-b)^2$.

To do:

We have to factorize the expression $16(a-b)^3-24(a-b)^2$.

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.

Here, we can factorize the expression $16(a-b)^3-24(a-b)^2$ by grouping similar terms and taking out the common factors. 

The terms in the given expression are $16(a-b)^3$ and $-24(a-b)^2$.

Therefore, by taking $(a-b)^2$ as common, we get,

$16(a-b)^3-24(a-b)^2=(a-b)^2[16(a-b)-24]$

Now, taking $8$ common in $[16(a-b)-24]$, we get,

$16(a-b)^3-24(a-b)^2=(a-b)^28[2(a-b)-3]$

$16(a-b)^3-24(a-b)^2=8(a-b)^2[2(a)-2(b)-3]$

$16(a-b)^3-24(a-b)^2=8(a-b)^2(2a-2b-3)$

Hence, the given expression can be factorized as $8(a-b)^2(2a-2b-3)$.

Akhileshwar Nani

Updated on: 06-Apr-2023

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