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bgbeidas
Not sure I fully understood your explanation to the question, is Bunuel around at all?


since thesee are standard number properties, even Bunuel will have to explain exactly what has been written ..
But may be I'll write it in detail..
\((\frac{2*\sqrt[3]{3}}{\sqrt{2}})^3\)

Substitute the values as following as-
\(2= \sqrt{2}*\sqrt{2}\)...
\(and 3= \sqrt[3]{3}*\sqrt[3]{3}*\sqrt[3]{3}\)...
\(2\sqrt{2}= \sqrt{2}*\sqrt{2}*\sqrt{2}\)...



=>\((\sqrt{2}*\sqrt[3]{3})^3\)
=> \(2\sqrt{2}*3 = 6\sqrt{2}\)
C
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bgbeidas
Apologies in advance for the poor formatting, it's my first time posting a question on here.

Which of the following is equal to (2^3√3/√2)^3?

A) 3√2/2
B) 3√2
C) 6√2
D) 12
E) 12√2

If anyone looking to edit the question can't interpret the above please let me know.


Hi,

i believe the Q is meant to be..
\((\frac{2*\sqrt[3]{3}}{\sqrt{2}})^3\)
=>\((\sqrt{2}*\sqrt[3]{3})^3\)
=> \(2\sqrt{2}*3 = 6\sqrt{2}\)
C
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Not sure I fully understood your explanation to the question, is Bunuel around at all?
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bgbeidas
Not sure I fully understood your explanation to the question, is Bunuel around at all?


since thesee are standard number properties, even Bunuel will have to explain exactly what has been written ..
But may be I'll write it in detail..
\((\frac{2*\sqrt[3]{3}}{\sqrt{2}})^3\)

Substitute the values as following as-
\(2= \sqrt{2}*\sqrt{2}\)...
\(and 3= \sqrt[3]{3}*\sqrt[3]{3}*\sqrt[3]{3}\)...
\(2\sqrt{2}= \sqrt{2}*\sqrt{2}*\sqrt{2}\)...



=>\((\sqrt{2}*\sqrt[3]{3})^3\)
=> \(2\sqrt{2}*3 = 6\sqrt{2}\)
C


I understand how you got 3, and how you should gt 2\sqrt{2} at the bottom.
But the 2 in the top part, when I cube it I get 8, not sure where that disappears in your process.
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Abhishek009
bgbeidas

=>\((\sqrt{2}*\sqrt[3]{3})^3\)
=> \(2\sqrt{2}*3 = 6\sqrt{2}\)
C

I understand how you got 3, and how you should gt 2\sqrt{2} at the bottom.
But the 2 in the top part, when I cube it I get 8, not sure where that disappears in your process.


I think u are clear with this part =>\((\sqrt{2}*\sqrt[3]{3})^3\)

=>\((\sqrt{2}*\sqrt[3]{3})^3\)

=> (\(2^{1/2}\) * \(3^{1/3}\))\(^{3}\)

=> \(2^{3/2}\)* \(3^{3/3}\)

=>\(2^1\)* \(2^{1/2}\)* \(3^1\) { Because \(\frac{3}{2}\) = \(1 \frac{1}{2}\) and \(\frac{3}{3}\) = 1 }

=> \(2^1\)* \(2^{1/2}\)* \(3^1\)

=> 2*\(\sqrt{2}\) * 3

=>6\(\sqrt{2}\)


Hope this helps !!
:-D :-D

Thank you Abhishek, you made the math much more clear. However, I am still confused as to how the root 2 jumped to the top, and where the original 2 up top disappeared to.
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bgbeidas
Thank you Abhishek, you made the math much more clear. However, I am still confused as to how the root 2 jumped to the top, and where the original 2 up top disappeared to.

I think you are referring to the the following part as under -

Quote:
\(\frac{2}{\sqrt{2}}\)

If that is the case then consider it this way -

\(\frac{2}{\sqrt{2}}\)

=>\((2 * \sqrt{2})/(\sqrt{2}* \sqrt{2})\)

=> \(2 * \sqrt{2}/2\)

=>\(\sqrt{2}\)

Hope this helps !! :-D :-D :-D
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bgbeidas
Abhishek009
bgbeidas

=>\((\sqrt{2}*\sqrt[3]{3})^3\)
=> \(2\sqrt{2}*3 = 6\sqrt{2}\)
C

I understand how you got 3, and how you should gt 2\sqrt{2} at the bottom.
But the 2 in the top part, when I cube it I get 8, not sure where that disappears in your process.


I think u are clear with this part =>\((\sqrt{2}*\sqrt[3]{3})^3\)

=>\((\sqrt{2}*\sqrt[3]{3})^3\)

=> (\(2^{1/2}\) * \(3^{1/3}\))\(^{3}\)

=> \(2^{3/2}\)* \(3^{3/3}\)

=>\(2^1\)* \(2^{1/2}\)* \(3^1\) { Because \(\frac{3}{2}\) = \(1 \frac{1}{2}\) and \(\frac{3}{3}\) = 1 }

=> \(2^1\)* \(2^{1/2}\)* \(3^1\)

=> 2*\(\sqrt{2}\) * 3

=>6\(\sqrt{2}\)


Hope this helps !!
:-D :-D

Thank you Abhishek, you made the math much more clear. However, I am still confused as to how the root 2 jumped to the top, and where the original 2 up top disappeared to.

ITS called " RATIONALISING THE DENOMINATOR"
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Fantastic reply Abhishek009 and chetan2u, many thanks.

In my opinion, the key part here is writting everything as exponents and then resolving the ^3 equation.

Thanks once again
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Abishek thank you! Super helpful. Could you clarify from the beginning how you got rid of the denominator listed in the question stem? It would be great if you could note why you did what you did next to the more complex lines.

Also, if you have any other lists of questions for roots, please share. Thanks!
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Never mind. I thought the original question was 2^3 instead of 3 root of 3...
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bgbeidas
Which of the following is equal to \((\frac{2*\sqrt[3]{3}}{\sqrt{2}})^3\)?

A) 3√2/2
B) 3√2
C) 6√2
D) 12
E) 12√2

We can simplify the expression inside the parentheses first.

Recall that x/√x = √x, so 2/√2 = √2 and hence 2 x (^3√3)/√2 = √2 x (^3√3)

Now we raise each term to the 3rd power:

[√2 x (^3√3)]^3 = (√2)^3 x (^3√3)^3 = 2√2 x 3 = 6√2

(Note: When we raise each factor to the 3rd power, we use the fact that:

(√x)^3 = x√x and (^3√x)^3 = x) .

Answer: C
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Why does the original equation have to be simplified?

Why can't tip 6 from "Exponents and roots: Tips and hints" post by bunuel be applied in this situation? (Can't post links yet)

I understand that you can't, because if you distribute the exponent and raise all the individual elements to 3 you get 8*3 / 2sqrt2 and ultimately 12/sqrt2.

What rule am I violating?
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Hi everyone,

Today I encounter this question on the one of the test on exam pack 2, and I cannot arrive at the OA of the question. Could anyone please help me solve it step by step?

Which of the following is equal to ( \(2^3\)\sqrt{\sqrt{3}}/2)\(3\)

A. \frac{3\sqrt{2}}{2}
B.3\sqrt{2}
C.6\sqrt{2}
D.12
E.12\sqrt{2}

OA is C
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NealC
Hi everyone,

Today I encounter this question on the one of the test on exam pack 2, and I cannot arrive at the OA of the question. Could anyone please help me solve it step by step?

Which of the following is equal to ( \(2^3\)\sqrt{\sqrt{3}}/2)\(3\)

A. \frac{3\sqrt{2}}{2}
B.3\sqrt{2}
C.6\sqrt{2}
D.12
E.12\sqrt{2}

OA is C

Merging topics.

Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html
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AfricanPrincess
The way the question has been asked is ambiguous because of what looks like 2^3 in the numerator.

Posted from my mobile device

Totally agreed with you, I encountered this question today, and thought the number 3 in this case is the power of 2 and therefore can not arrived at the correct answer. I had to guess after 4m trying hopelessly to solve it, very frustrating as it is just the second question, I thought the answer is wrong. I think GMAT should make it clear which number belong to which.
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Bunuel
NealC
Hi everyone,

Today I encounter this question on the one of the test on exam pack 2, and I cannot arrive at the OA of the question. Could anyone please help me solve it step by step?

Which of the following is equal to ( \(2^3\)\sqrt{\sqrt{3}}/2)\(3\)

A. \frac{3\sqrt{2}}{2}
B.3\sqrt{2}
C.6\sqrt{2}
D.12
E.12\sqrt{2}

OA is C

Merging topics.

Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html

Sorry Bunuel,

I had tried to search it in the forum and cannot find the question. Therefore I though no one encounter it before and posted it. The way this question is worded made me very frustrated, I really want to look for its discussion, and thus when I couldn't find it, I immediately posted it. My apology, I will search harder the next time before attempting to post any question.
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IMO - the wording on this problem is ambiguous. How am I supposed to know it is 2(CUBERT(3)) instead of 2^3(Sqrt(3))???
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