Question

Simplify a + b + C whole square + a minus b + C whole square

Answer 1

(a+b+c)²+(a-b+c)²
=(a)² + (b)² +(c)²+2ab +2bc +2ac+{(a)²+(b)²+(c)²+2ac-2bc-2ab}
=2(a)²+2(b)²+2(c)² +2ab +2bc+2ac-2ab+2ac-2bc
=2(a)²+2(b)²+2(c)²+4ac

Answer 2

Answer: The simplification of $(a+b+c)^{2} -(a-b+c)^{2}$ is $4ab+bc$

Step-by-step explanation:

Given: We have given $(a+b+c)^{2} -(a-b+c)^{2}$

To find:We have to simplify  $(a+b+c)^{2} -(a-b+c)^{2}$

Explanation:

Step 1:Apply the identity $a^{2} -b^{2} = (a+b)(a-b)$

         where $a=(a+b+c)$ and $b=(a+b+c)$ respectively.

Step 2:On substituting the value of a and b on the above identity we get,

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

                                            $= (a+b+c+a-b+c)(a+b+c-a+b-c)$

                                            $= (2a+2c)(2b)$

                                            $= 4ab+4bc$

∴       $(a+b+c)^{2} -(a-b+c)^{2}$ $= 4ab+4bc$.

Concept:Algebraic Identities are algebraic equation which are always true for every value of the variable in them.The algebraic identities that are valid for all values of variables in them are called algebraic identities. It is used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and solving different polynomials. They contain variable and constant on both the side of polynomial i.e.$LHS$ and $RHS$. In algebraic identity, $LHS$ must be equal to$RHS$.

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