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

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GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
If x^3(x^2+y^2)=z^2, is xyz=0?  [#permalink]

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25 Feb 2019, 16:03
00:00

Difficulty:

85% (hard)

Question Stats:

35% (02:23) correct 65% (01:45) wrong based on 20 sessions

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GMATH practice exercise (Quant Class 14)

If $$\,{x^3}\left( {{x^2} + {y^2}} \right) = {z^2}\,$$, is $$\,xyz = 0\,$$ ?

$$\left( 1 \right)\,\,{y^3}\left( {{y^2} + {z^2}} \right) = {x^2}\,\,$$

$$\left( 2 \right)\,\,{z^3}\left( {{z^2} + {x^2}} \right) = {y^2}$$

_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
If x^3(x^2+y^2)=z^2, is xyz=0?  [#permalink]

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25 Feb 2019, 17:51
fskilnik wrote:
GMATH practice exercise (Quant Class 14)

If $$\,{x^3}\left( {{x^2} + {y^2}} \right) = {z^2}\,$$, is $$\,xyz = 0\,$$ ?

$$\left( 1 \right)\,\,{y^3}\left( {{y^2} + {z^2}} \right) = {x^2}\,\,$$

$$\left( 2 \right)\,\,{z^3}\left( {{z^2} + {x^2}} \right) = {y^2}$$

$${x^3}\left( {{x^2} + {y^2}} \right) = {z^2}\,\,\,\,\,\left( * \right)$$

$$xyz\,\,\mathop = \limits^? \,\,0$$

$$\left( {1 + 2} \right)\,\,\left\{ \matrix{ \,{y^3}\left( {{y^2} + {z^2}} \right) = {x^2}\,\,\left( * \right) \hfill \cr \,{z^3}\left( {{z^2} + {x^2}} \right) = {y^2}\,\,\left( * \right) \hfill \cr} \right.\,\,$$

$${\rm{Take}}\,\,\,\left\{ \matrix{ \,\left( {x;y;z} \right) = \left( {0;0;0} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr \,\left( {**} \right)\,\,\,\left( {x;y;z} \right) = \left( {{1 \over {\root 3 \of 2 }};{1 \over {\root 3 \of 2 }};{1 \over {\root 3 \of 2 }}} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.$$

$$\left( {**} \right)\,\,{\rm{Explore}}\,\,{\rm{symmetries(!),}}\,\,\underline {{\rm{trying}}} \,\,\,\left( {x,y,z} \right) = \left( {k,k,k} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{any}}\,\,\left( * \right)} \,\,\,\,{k^3}\left( {2{k^2}} \right) = {k^2}\,\,\,\,\mathop \Rightarrow \limits^{k\, \ne \,0} \,\,\,2{k^3} = 1\,\,\,\,\, \Rightarrow \,\,\,\,k = {1 \over {\root 3 \of 2 }}\,\,\,\,{\rm{viable}}!$$

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)
Our high-level "quant" preparation starts here: https://gmath.net
If x^3(x^2+y^2)=z^2, is xyz=0?   [#permalink] 25 Feb 2019, 17:51
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# If x^3(x^2+y^2)=z^2, is xyz=0?

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