Question

please solve (a+b-c)whole 2 and(4a+2b+c)whole 2

Answer 1

Hello friends!!

Here is your answer :

Using identity :

( x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx


(a)
${(a + b - c)}^{2}$


$= > {(a)}^{2} + {(b)}^{2} + {( - c)}^{2} + 2(a)(b) + 2(b)( - c) + 2( - c)(a)$


$= > {a}^{2} + {b}^{2} + {c}^{2} + 2ab - 2bc - ca$

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( b)
${(4a + 2b + c)}^{2}$


$= > {(4a)}^{2} + {(2b)}^{2} + {(c)}^{2} + 2(4a)(2b) + 2(2b)(c) + 2(c)(4a)$


$= > {16a}^{2} + {4b}^{2} + {c}^{2} + 16ab + 4bc + 8ca$

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Hope it helps you...

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