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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## If x^3 (x^2+y^2) = z^2 , is xyz = 0 ? tagged by: fskilnik@GMATH ##### This topic has 1 expert reply and 0 member replies ### GMAT/MBA Expert ## If x^3 (x^2+y^2) = z^2 , is xyz = 0 ? ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult GMATH practice exercise (Quant Class 14) If x^3 (x^2+y^2) = z^2 , is xyz = 0 ? (1) y^3 (y^2+z^2) = x^2 (2) z^3 (z^2+x^2) = y^2 Answer: ____(E)__ _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 fskilnik@GMATH wrote: GMATH practice exercise (Quant Class 14) If x^3 (x^2+y^2) = z^2 , is xyz = 0 ? (1) y^3 (y^2+z^2) = x^2 (2) z^3 (z^2+x^2) = y^2 $${x^3}\left( {{x^2} + {y^2}} \right) = {z^2}\,\,\,\,\,\left( * \right)$$ $$xyz \,\, \mathop = \limits^? \,\,0$$ $$\left( {1 + 2} \right)\,\,\left\{ \matrix{ \,{y^3}\left( {{y^2} + {z^2}} \right) = {x^2}\,\,\left( * \right) \hfill \cr \,{z^3}\left( {{z^2} + {x^2}} \right) = {y^2}\,\,\left( * \right) \hfill \cr} \right.\,\,$$ $${\rm{Take}}\,\,\,\left\{ \matrix{ \,\left( {x;y;z} \right) = \left( {0;0;0} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr \,\left( {**} \right)\,\,\,\left( {x;y;z} \right) = \left( {{1 \over {\root 3 \of 2 }};{1 \over {\root 3 \of 2 }};{1 \over {\root 3 \of 2 }}} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.$$ $$\left( {**} \right)\,\,{\rm{Explore}}\,\,{\rm{symmetries(!),}}\,\,\underline {{\rm{trying}}} \,\,\,\left( {x,y,z} \right) = \left( {k,k,k} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{any}}\,\,\left( * \right)} \,\,\,\,{k^3}\left( {2{k^2}} \right) = {k^2}\,\,\,\,\mathop \Rightarrow \limits^{k\, \ne \,0} \,\,\,2{k^3} = 1\,\,\,\,\, \Rightarrow \,\,\,\,k = {1 \over {\root 3 \of 2 }}\,\,\,\,{\rm{viable}}!$$ The correct answer is (E). We follow the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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