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# M26-36

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Math Expert
Joined: 02 Sep 2009
Posts: 52228

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16 Sep 2014, 00:26
1
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Difficulty:

95% (hard)

Question Stats:

32% (01:02) correct 68% (01:15) wrong based on 225 sessions

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If $$x$$, $$y$$ and $$z$$ are non-zero numbers, what is the value of $$\frac{x^3+y^3+z^3}{xyz}$$?

(1) $$xyz=-6$$

(2) $$x+y+z=0$$

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Math Expert
Joined: 02 Sep 2009
Posts: 52228

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16 Sep 2014, 00:26
3
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Official Solution:

(1) $$xyz=-6$$. Infinitely many combinations of $$x$$, $$y$$ and $$z$$ are possible which will give different values of the expression in the stem: try $$x=y=1$$ and $$y=-6$$ or $$x=1$$, $$y=2$$, $$z=-3$$. Not sufficient.

(2) $$x+y+z=0$$. Rearrange: $$x=-(y+z)$$ and substitute this value of $$x$$ into the expression in the stem to get $$\frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}$$, as $$x=-(y+z)$$ then:

$$\frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3$$. Sufficient.

Must know for the GMAT:$$(x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3$$ and $$(x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3$$.

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Joined: 16 Jul 2014
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04 Nov 2014, 17:21
1
Once it gets to this stage:

−3yz(y+z)
xyz

yz can be canceled out so we're only left with:

−3(y+z)
x

And since x = -(y+z)

3x
x

= 3
Manager
Joined: 17 May 2017
Posts: 130
GPA: 3

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30 Aug 2017, 00:43
1
Can I expect such type of questions in GMAT
this is more or less like CAT(Entrance Exam for Indian Institute of Management) Question
Manager
Joined: 20 Feb 2017
Posts: 164
Location: India
Concentration: Operations, Strategy
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17 Dec 2017, 12:29
3
1
THERE IS A SIMPLE PROPERTY
IF X+Y+Z = 0
THEN X^3+Y^3+Z^3 = 3XYZ
HOPE IT HELPS
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Manager
Joined: 30 May 2017
Posts: 135
Location: United States
Schools: HBS '21
GMAT 1: 690 Q50 V32
GRE 1: Q168 V164
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05 May 2018, 17:46
THERE IS A SIMPLE PROPERTY
IF X+Y+Z = 0
THEN X^3+Y^3+Z^3 = 3XYZ
HOPE IT HELPS

I assume we only expect the property above when NONE of X,Y and Z is equal to zero.
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Kindly press the +1Kudos if you like the explanation. Thanks a lot!!!

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Joined: 08 Jun 2015
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Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
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24 Jul 2018, 05:19
Use the property -> x^3+y^3+z^3 - 3xyz=(x+y+z)*(x^2+y^2+z^2-xy-yz-zx) . If x+y+z=0 , The required expression comes to 3. I doubt though if such a formulaic question will ever come on the GMAT !!!
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Manager
Joined: 20 Jul 2018
Posts: 89
GPA: 2.87

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05 Sep 2018, 08:23
statement 1 not sufficient as many combinations of x,y and z are possible which can result into different values.
statement 2 implies:
$$x+y+z=0$$
$$x=-(y+z)$$
substitute for x
$$\frac{(-y^3 - z^3 - 3y^2z -3yz^z +y^3 + z^3)}{xyz}$$
$$\frac{-3yz(y+z)}{xyz}$$
$$\frac{-3yz(y+z)}{-yz(y+z)}$$
$$3$$
Sufficient

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Hasnain Afzal

"When you wanna succeed as bad as you wanna breathe, then you will be successful." -Eric Thomas

Intern
Joined: 31 Jul 2017
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06 Sep 2018, 20:05
Any OG/GMAT question which employs (x+y)^3 formula in its solution?
Manager
Joined: 26 Sep 2018
Posts: 57
Location: Sweden

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11 Jan 2019, 11:20
Bunnel you are some legend
Re: M26-36 &nbs [#permalink] 11 Jan 2019, 11:20
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# M26-36

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