A simple but painful calculation.

Question

$$S=(x_3-x_1)^2+(x_4-x_2)^2+(x_5-x_3)^2+(x_1-x_4)^2+(x_2-x_5)^2$$.

$$S'=(x_3-x_1)^2+(x_4+x_2)^2+(x_5-x_2-x_3)^2+(x_1+x_2-x_4)^2+(-x_2-x_5)^2$$

Find $S$–$S'$?

I thought that, I should manipulate the above equations on my own but I got stuck, How can I reach to the answer???

Answer 1

Hint: Evaluate the difference term by term, i.e. square by square, and use the fact that $$a^2-b^2=(a-b)(a+b).$$ There is nothing difficult or painful about this.

In fact, it is quickly clear that in each case you get a multiple of $x_2$. The expression as a whole is quadratic, so it is a linear factor times $x_2$.

Answer 2

If you don't want to perform the expansion manually, may be you could use Symbola or a similar software.

$$S-S'=(x_3-x_1)^2+(x_4-x_2)^2+(x_5-x_3)^2+(x_1-x_4)^2+(x_2-x_5)^2 -((x_3-x_1)^2+(x_4+x_2)^2+(x_5-x_2-x_3)^2+(x_1+x_2-x_4)^2+(-x_2-x_5)^2)$$

is

$$-2x_{3}x_{2}-2x_{2}^2-2x_{4}x_{2}-2x_{1}x_{2}-2x_{2}x_{5}$$