Elementary algebraic equivalent expression for $(y_1-y_2)(y_2-y_3)(y_3-y_1)$.

Question

*It is a question unanswered in my earlier post at: https://math.stackexchange.com/a/2755962/424260. *

I have been given that: $(y_1-y_2)(y_2-y_3)(y_3-y_1) = y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1)$.

But, the derivation below shows that there is reversal of terms inside each of the $3$ brackets.
$$(y_1-y_2)(y_2-y_3)(y_3-y_1) = (y_1y_2-y_1y_3-y_2^2+y_2y_3)(y_3-y_1)$$ $$=-y_1y_3^2 -y_2^2y_3 +y_2y_3^2-y_1^2y_2+y_1^2y_3+y_1y_2^2$$ $$=y_1y_3(y_1-y_3)+y_2y_3(y_3-y_2)+y_1y_2(y_2-y_1)$$

Kindly help me with finding error in my derivation.

Answer 1

The statement you have been given is missing a minus sign on one end.

Your deductions are absolutely correct. Notice that:

$$y_1y_2(y_1-y_2)=-y_1y_2(y_2-y_1)$$ and the same holds with the other two sets.

Answer 2

The given statement $$(y_1-y_2)(y_2-y_3)(y_3-y_1) = y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1)$$ is not correct. Just check it for $$ y_1=1, y_2=2, y_3 =3$$ and you get $2=-2$

The correct version is $$(y_1-y_2)(y_2-y_3)(y_3-y_1)=y_1y_3(y_1-y_3)+y_2y_3(y_3-y_2)+y_1y_2(y_2-y_1)$$ which you have obtained.