Question:
Find the value
$(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c) \times(b-c+a)$
Solution:
Let (a - b + c) = x and (b - c + a) = y
$=x^{2}+y^{2}+2 x y$
Using the identity $(a+b)^{2}=a^{2}+b^{2}+2 a b$
$=(x+y)^{2}$
Now, substituting x and y
$(a-b+c+b-c+a)^{2}$
Cancelling -b, +b & + c, -c
$=(2 a)^{2}$
$=4 a^{2}$
$\therefore(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c) \times(b-c+a)=4 a^{2}$
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