Fast algebraic expansion Is there an algebraic trick to expand the following expression without multiplying each term with another, expanding the standard way it gives $6+6+6=18$ terms and then cancelling the same terms with opposite signs to get the final result, is there any shortcut?
$(b+c-a)(y+z)+(c+a-b)(z+x)+(a+b-c)(x+y)$
 A: Arrange the three terms in such way:
$$ (-a + b + c) (0 + y + z) +$$
$$ (+a - b + c) (x + 0 + z) +$$
$$ (+a + b - c) (x + y + 0) =$$
(I had to use the equal sign to line them up)
Let's look at the $a$ column. $y$ and $z$ are subtracted away leaving two $x$'s in the result. Therefore, we have $2ax$ from the $a$ column. Similarly, we get $2by$ and $2cz$. Hence $2(ax+by+cz)$ is the answer.
A: There is a cyclic symmetry in the expression.  The second term is obtained from the first by replacing $a \to b, b \to c, c \to a, x \to y, y \to z, z \to x$.  The third term is obtained from the second in the same way.  And if you apply the symmetry to the third term, you get the first term again.
There are at most 9 terms in the expanded form: each of a,b,c multiplied by each of x,y,z.  But the symmetry means that the coefficient of $ax$ will be the same as the coefficients of $by$ and $cz$, and similarly for all the other terms.
Now take any particular term that occurs in the expansion of the first term, say $by$.  You notice that $by$ also occurs in the third term as a positive term, but not in the second term at all.  So $by$ will have a coefficient 2 in the final answer.  By applying the cyclic symmetry, the terms $ax$ and $cz$ also have coefficient 2.
Now take one of the other terms in the expansion of the first term, say $bz$.  The second term has a $-bz$ and the third term, no $bz$.  So there is no $bz$ and by symmetry no $cx$ nor $ay$.
Similarly, check $bx$.  It cancels out of the second and third term.  Therefore, by symmetry, so does $cy$ and $az$.
So final answer: $2(ax+by+cz)$.
