Last visit was: 15 Jul 2025, 05:26 It is currently 15 Jul 2025, 05:26
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Jul 2025
Posts: 102,576
Own Kudos:
Given Kudos: 98,190
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,576
Kudos: 741,535
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
EnriqueDandolo
Joined: 07 Aug 2024
Last visit: 13 Jul 2025
Posts: 18
Own Kudos:
Given Kudos: 18
Posts: 18
Kudos: 1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Jul 2025
Posts: 102,576
Own Kudos:
741,535
 [1]
Given Kudos: 98,190
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,576
Kudos: 741,535
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
EnriqueDandolo
Joined: 07 Aug 2024
Last visit: 13 Jul 2025
Posts: 18
Own Kudos:
Given Kudos: 18
Posts: 18
Kudos: 1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hey Bunuel, thanks for your help, what you mean is that in this problem if I were to cube root all terms in the numerator and denominator, then I would have the following fraction?

Cube root(x^3 + y^3 + z^3)/Cube root(xyz)

Thanks in advance for your help.

Bunuel
EnriqueDandolo
Hi experts Bunuel KarishmaB, I was wondering if you could please help me point what I am doing wrong here. To evaluate statement 2, I would actually do the following to the stimulus:

Cube root all terms and numbers in the numerator and denominator so that the expression turns into the following:

Cube root(x^3) + Cube root(y^3) +Cube root(z^3) / Cube root(xyz)
*I did this because even though I can't take the square root of a variable with an even exponent because I don't know if the result will be positive or negative. Since we are working here with an odd exponent, i figured that there will only be 1 case, either the result is positive or negative.

After simplifying, the expression would turn into:
x+y+z/cube root(xyz).

Since the second statement tells me that x+y+z=0, then the result will be:
0/cube root(xyz)

Thus, the result of the expression will always be 0.

Many thanks in advance for your help.

Bunuel
9. 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) \(x+y+z=0\)

(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) \(x+y+z=0\) --> \(x=-(y+z)\) --> substitute this value of x into the expression in the stem --> \(\frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}\), 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 \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).

Answer: B.

The point is \(\sqrt{x^3+y^3+z^3}\) does not equal to x^3+y^3+z^3. You'd be right if it were (x + y + x)^3 instead.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Jul 2025
Posts: 102,576
Own Kudos:
Given Kudos: 98,190
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,576
Kudos: 741,535
Kudos
Add Kudos
Bookmarks
Bookmark this Post
EnriqueDandolo
Hey Bunuel, thanks for your help, what you mean is that in this problem if I were to cube root all terms in the numerator and denominator, then I would have the following fraction?

Cube root(x^3 + y^3 + z^3)/Cube root(xyz)

Thanks in advance for your help.

_______________
Right.
   1   2   3 
Moderator:
Math Expert
102576 posts