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Show that the expression $$(x^2-yz)^3+(y^2-zx)^3+(z^2-xy)^3-3(x^2-yz)(y^2-zx)(z^2-xy)$$ is a perfect square and Find its square root.

Vidyanshu Mishra
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Lelouch vi Britannia
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  • The effort put into this type of question should not really have been upvoted. As such I am downvoting it. –  Dec 08 '16 at 12:17
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    COULD WE GET YOU TO STOP USING UPPERCASE AND TO FORMULATE A MEANINGFUL QUESTION INSTEAD OF THE CURRENT ONE WHICH IS SO GENERIC THAT IT IS LARGELY MEANINGLESS? MOREOVER, IS THIS A HOMEWORK PROBLEM? AND WHAT HAVE YOU TRIED IN ORDER TO OBTAIN A SOLUTION YOURSELF? – Hans Hüttel Dec 08 '16 at 12:19
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    @HansHüttel You made your point. In general, you don't stop a fire by setting another fire. Could I get you to stop using uppercase and formulate a meaningful comment instead of the current one which is so juvenile that it becomes meaningless?: E.g. "You're an ahole"! ... "No, you're the ahole!". I.e., How effective is it to stop someone from shouting, by shouting back? Why is it okay for you to shout at someone who is shouting? Sorry for the old adage, but, "two wrongs don't make a right" – amWhy Dec 08 '16 at 19:48

2 Answers2

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The given expression can be factored as $$[(x+y+z)(x^2+y^2+z^2-xy-yz-zx)]^2.$$

Robert Z
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$$(x^2 - yz)^3 + (y^2 - zx)^3 + (z^2 - xy)^3 - 3 (x^2 - yz) (y^2 - zx) (z^2 - xy) =(x^3 + y^3 - 3 x y z + z^3)^2$$

King Ghidorah
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  • How do you know that is the square root to choose, why not $3xyz-x^3-y^3-z^3$? –  Dec 08 '16 at 13:36
  • @Bacon $$(3xyz-x^3-y^3-z^3)^2 = (- (3xyz-x^3-y^3-z^3))^2 = (x^3 + y^3 - 3 x y z + z^3)^2$$ – amWhy Dec 08 '16 at 19:28
  • @amWhy Trivial, I suppose, Thanks for that, just wanted to be clear, after all there are two solutions are there not? –  Dec 09 '16 at 09:12