Find the value

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

$=(2 a)^{2}$

$=4 a^{2}$

$\therefore(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c) \times(b-c+a)=4 a^{2}$