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If x^3 (x^2+y^2) = z^2 , is xyz = 0 ?

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If x^3 (x^2+y^2) = z^2 , is xyz = 0 ?

by fskilnik@GMATH » Mon Feb 25, 2019 2:53 pm

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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: [spoiler]____(E)__[/spoiler]
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Source: — Data Sufficiency |
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If x^3 (x^2+y^2) = z^2 , is xyz = 0 ?

by fskilnik@GMATH » Mon Feb 25, 2019 4:35 pm

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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
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