Question
Factorize:
$(a - b + c)^2 + (b - c + a)^2 + 2(a - b + c)(b - c + a)$
$(a - b + c)^2 + (b - c + a)^2 + 2(a - b + c)(b - c + a)$
✓
Answer
$(a - b + c)^2 + (b - c + a)^2 + 2(a - b + c)(b - c + a)$
Let $(a - b + c) = x$ and $(b - c + a) = y$
$= x^2 + y^2 + 2xy$
Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$
$= (x + y)^2$
Now, substituting $x$ and $y$
$(a - b + c + b - c + a)^2$
Cancelling $-b, +b + c, -c$
$= (2a)^2$
$= 4a^2$
$\therefore (a - b + c)^2 + (b - c + a)^2 + 2(a - b + c)(b - c + a)$
$ = 4a^2$
Let $(a - b + c) = x$ and $(b - c + a) = y$
$= x^2 + y^2 + 2xy$
Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$
$= (x + y)^2$
Now, substituting $x$ and $y$
$(a - b + c + b - c + a)^2$
Cancelling $-b, +b + c, -c$
$= (2a)^2$
$= 4a^2$
$\therefore (a - b + c)^2 + (b - c + a)^2 + 2(a - b + c)(b - c + a)$
$ = 4a^2$
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