Question

prove that 2a^2-2b^2c^2 -2ab-2bc-2ca=(a-b)^2+(b-c) ^2+(c-a) ^2

Answer 1

We have $RHS=(a-b)^2+(b-c)^2+(c-a)^2$. Expanding each square,

Using the formula $(x-y)^2=x^2+y^2-2xy$, we get

$RHS=a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca$.

Rearranging the above expression,

[tex]RHS=a^2+b^2+b^2+c^2+c^2+a^2-2ab-2bc-2ca\\ RHS=2a^2+2b^2+2c^2-2ab-2bc-2ca\\ RHS=LHS[/tex].

The proof is complete.