Question

Factorize (a – b + c)2 + (b – c + a) 2 + 2(a – b + c) (b – c + a)​

Answer 1

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(a – b + c)^2 + ( b – c + a)^2 + 2(a – b + c) (b – c + a)

{Because p^2 + q^2 + 2pq = (p + q)^2}

Here p = a – b + c and q = b – c + a

= [a – b + c + b- c + a]^2

= (2a)^2

= 4a^2

Answer 2

Answer:

4a²

Step-by-step explanation:

(a-b+c)²+(b-c+a)²+2(a-b+c)(b-c+a)

= [(a-b+c)+(b-c+a)]²

/* By algebraic identity:

x²+y²+2xy = (x+y)² */

= (a-b+c+b-c+a)²

= (2a)²

= 4a²

Therefore,

(a-b+c)²+(b-c+a)²+2(a-b+c)(b-c+a)= 4a²

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