<s>[INST] Prove that $a(x+y+z) = x(a+b+c)$ If $(a^2+b^2 +c^2)(x^2+y^2 +z^2) = (ax+by+cz)^2$
Then prove that $a(x+y+z) = x(a+b+c)$
I did expansion on both sides and got:
$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2(abxy+bcyz+cazx) $ 
but can't see any way to prove $a(x+y+z) = x(a+b+c)$. How should I proceed?
 [/INST] $$a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2byxz$$
$$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2axcz+2byxz$$
$$a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2byxz+c^2y^2=0$$
$$(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0$$
From here the answer is clear. All three terms are zero.
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