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555-605 (Medium)|   Algebra|   Exponents|   Roots|      
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Bunuel
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NEW!!! Tough and tricky exponents and roots questions 08 Mar 2025, 15:05
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EnriqueDandolo
Hey Bunuel, I was wondering if you could please explain step by step in more detail how did you apply the algebraic identity (x^2 - y^2)=(x+y)(x-y) to the fraction

Square root(2)*[Square root(x) + Square root(y)] /(x-y).

Many thanks in advance.

I am confused because after applying the same algebraic identity I got as a result the following:

Square root(2)*[Square root(x) + Square root(y)]*(x+y)/(x^2 - y^2)

Bunuel
3. If \(x>y>0\) then what is the value of \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)?
(1) \(x+y=4+2\sqrt{xy}\)
(2) \(x-y=9\)

\(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\) --> factor out \(\sqrt{2}\) from the nominator and apply \(a^2-b^2=(a-b)(a+b)\) to the expression in the denominator: \(\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}\). So we should find the value of \(\sqrt{x}-\sqrt{y}\).

(1) \(x+y=4+2\sqrt{xy}\) --> \(x-2\sqrt{xy}+y=4\) --> \((\sqrt{x}-\sqrt{y})^2=4\) --> \(\sqrt{x}-\sqrt{y}=2\) (note that since \(x-y>0\) then the second solution \(\sqrt{x}-\sqrt{y}=-2\) is not valid). Sufficient.

(2) \(x-y=9\). Not sufficient.

Answer: A.
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We are applying \(a^2-b^2=(a-b)(a+b)\) to the expression in the denominator"

In the denominator we have x - y, which can be written as \(x-y=(\sqrt{x})^2-(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})\).

Hope it's clear.
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NEW!!! Tough and tricky exponents and roots questions 09 Mar 2025, 11:05
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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.
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NEW!!! Tough and tricky exponents and roots questions 09 Mar 2025, 11:32
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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.
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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.
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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.
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Bunuel
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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.
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_______________
Right.
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AditiDeokar
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NEW!!! Tough and tricky exponents and roots questions 24 Dec 2025, 11:41
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Hi Bunuel, are these kind of D/S questions asked in the GMAT FE, as I RECALL READING PURE ALGEBRA Q's are not asked in DS
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Bunuel
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NEW!!! Tough and tricky exponents and roots questions 24 Dec 2025, 12:37
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AditiDeokar
Hi Bunuel, are these kind of D/S questions asked in the GMAT FE, as I RECALL READING PURE ALGEBRA Q's are not asked in DS
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No, such type of DS questions are no longer tested on the exam.
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