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555-605 (Medium)|   Arithmetic|   Roots|      
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Bunuel
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If x = 216 and y = 125, what is the value of x2/3 + y2/3 + 2(xy)1/3 + 06 Apr 2021, 07:45
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E
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35% (medium)

Question Stats:

74% (02:11) correct 26% (02:31) wrong based on 609 sessions

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If x = 216 and y = 125, what is the value of \(\sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} + 2(xy)^{\frac{1}{3}}} - \sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} − 2(xy)^{\frac{1}{3}}}\)?

(A) 0
(B) 1
(C) 4
(D) 5
(E) 10
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E
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Ravi1410
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If x = 216 and y = 125, what is the value of x2/3 + y2/3 + 2(xy)1/3 + 06 Apr 2021, 08:09
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216=2^3*3^3 and 125=5^3. According to the equation x^(2/3) + y^(2/3) = 36 + 25 and 2(x*y)^(1/3) = 60.

In the first half, rootof(36+25+60) = rootof(121) = 11
In the second half, rootof(36+25-60) = rootof(1) = 1.

Adding the both gives us 12. But there is no option. I guess the question should be 11-1 = 10 as we have that option.
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If x = 216 and y = 125, what is the value of x2/3 + y2/3 + 2(xy)1/3 + 06 Apr 2021, 08:43
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Ans should be 12 (11+1)
But since it’s not there, it might be a mistake in question
Hence Marking E (10= 11-1)

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If x = 216 and y = 125, what is the value of x2/3 + y2/3 + 2(xy)1/3 + 09 Apr 2021, 01:50
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Ravi1410
216=2^3*3^3 and 125=5^3. According to the equation x^(2/3) + y^(2/3) = 36 + 25 and 2(x*y)^(1/3) = 60.

In the first half, rootof(36+25+60) = rootof(121) = 11
In the second half, rootof(36+25-60) = rootof(1) = 1.

Adding the both gives us 12. But there is no option. I guess the question should be 11-1 = 10 as we have that option.
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Fixed the typo. Thank you.
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If x = 216 and y = 125, what is the value of x2/3 + y2/3 + 2(xy)1/3 + 22 Nov 2024, 01:09
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Is there a simpler way to do this since I'm unable to understand the explanation given.
Bunuel
If x = 216 and y = 125, what is the value of \(\sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} + 2(xy)^{\frac{1}{3}}} - \sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} − 2(xy)^{\frac{1}{3}}}\)?

(A) 0
(B) 1
(C) 4
(D) 5
(E) 10
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If x = 216 and y = 125, what is the value of x2/3 + y2/3 + 2(xy)1/3 + 05 Apr 2025, 17:23
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Mayzone
Is there a simpler way to do this since I'm unable to understand the explanation given.
Bunuel
If x = 216 and y = 125, what is the value of \(\sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} + 2(xy)^{\frac{1}{3}}} - \sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} − 2(xy)^{\frac{1}{3}}}\)?

(A) 0
(B) 1
(C) 4
(D) 5
(E) 10
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The above equation can be re-written as |x^1/3 + y^1/3| - |x^1/3 - y^1/3| -----(1)
using first the algebraic identity (a+b)^2= a^2 + 2ab +b^2 and (a-b)^2= a^2 - 2ab +b^2

Second concept, root(x^2) = |x|

Since the number x and y are postive it's cube root will also be postive.

|216^1/3 + 125^1/3| - |216^1/3 - 125^1/3|
|6+5|-|6-5|
11-1
=10. Hence, E.
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If x = 216 and y = 125, what is the value of x2/3 + y2/3 + 2(xy)1/3 + 12 Jul 2025, 22:21
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remove the under root because you can re-write is as (a+b)sq + (a-b)sq
where a=x raised to 1/3 and b=y raised to 1/3
you get 2(x raise to 1/3) when you solve which is 2(125 to 1/3) which is 2(5) which eqauls 10

so 10 is the answer
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