\frac{(a-b+c)^2-(a-b-c)^2}{b-a} is equal to

CTET 2 Math and Science 07 Jul 2024 Paper

Question : 36 of 150

Marks: +1, -0

\;{(a-b+c)^2-(a-b-c)^2}/{b-a}

is equal to

Solution:

a^2-b^2=(a-b)(a+b)

Explanation
We have

\;{(a-b+c)^2-(a-b-c)^2}/{b-a}

\;=\;{(a-b+c-(a-b-c))(a-b+c+(a-b-c))}/{b-a}

\;=\;{(a-b+c-a+b+c)(a-b+c+a-b-c)}/{b-a}

\;=\;{(c+c)(a-b+a-b)}/{b-a}

\;=\;{(2 c)(2 a-2 b)}/{b-a}=\;{(2 c) ×-2(b-a)}/{b-a}=-4 c

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