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# If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?

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Math Expert
Joined: 02 Sep 2009
Posts: 58445
If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?  [#permalink]

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06 Nov 2014, 09:02
5
15
00:00

Difficulty:

95% (hard)

Question Stats:

39% (02:03) correct 61% (02:10) wrong based on 466 sessions

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Tough and Tricky questions: Algebra.

If $$a^{\frac{2}{3}} - b^{\frac{2}{3}} = 12$$, then $$\sqrt[3]{a} + \sqrt[3]{b} =$$ ?

(1) $$\sqrt[3]{a} = \sqrt[3]{b} + 2$$

(2) a = 64

Kudos for a correct solution.

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Manager
Joined: 21 Jan 2014
Posts: 61
WE: General Management (Non-Profit and Government)
Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?  [#permalink]

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10 Nov 2014, 05:55
4
1
As per question : a^(2/3)-b^(2/3)=12........(1)

so we can break this in terms of a^2-b^2=(a+b)(a-b)
(a^1/3)^2-(b^1/3)^2 = (a^1/3 +b^1/3)(a^1/3 - b^1/3)=12 ......(2)

1) a^1/3 = b^1/3 +2
a^1/3 - b^1/3 =2
So inserting this value in equation (2) will solve this puzzle. SUFFICIENT

2) a=64
taking this value in equation (1)
64^2/3 - b^2/3=12
16 -12 =b^2/3
so b^2/3=4 and b^1/3=+-2.
this two different values for b^1/3 will create new puzzle rather solving it.INSUFFICIENT

##### General Discussion
Intern
Joined: 27 Jul 2012
Posts: 25
Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?  [#permalink]

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06 Nov 2014, 13:39
Both are sufficient

statement 1: use a^2-b^2=(a+b)(a-b). Tis gives us answer 6.
Sufficient

Statement 2
Put 64 in first equation. that gives us b=8.
Put both a and b into final question and find answer as 6.
Sufficient
Intern
Joined: 19 Oct 2014
Posts: 8
Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?  [#permalink]

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08 Nov 2014, 23:14
1
Given:
$$a^{2/3}-b^{2/3}= (a^{1/3}+b^{1/3})*(a^{1/3}-b^{1/3}) = 12$$

$$a^{1/3}+b^{1/3} = ?$$

Statement 1:

$$(a^{1/3}-b^{1/3})=2$$

Therefore $$a^{1/3}+b^{1/3} = 6$$

Sufficient.

Statement 2:

a=64

This can be used to solve for b and hence is sufficient.

Kudos please! Mr Bunuel, the quant king!!
Math Expert
Joined: 02 Sep 2009
Posts: 58445
Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?  [#permalink]

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10 Nov 2014, 05:35
D is not correct. Anyone else wants to try?
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Intern
Joined: 07 Mar 2014
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Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?  [#permalink]

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14 Apr 2015, 04:16
for a=64
the sol for b = +/- 8. This gives two values of B, therefore insufficient.
Senior Manager
Joined: 02 Apr 2014
Posts: 468
Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34
GPA: 3.5
Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?  [#permalink]

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08 Nov 2017, 14:00
2
1

$$a ^ {2/3} - b ^ {2/3} = 12$$

Let $$x = a ^ {1/3}, y = b ^ {1/3}$$ => $$x ^ 2 - y ^ 2 = 12$$ =>$$(x + y)(x - y)$$ = 12

Question :$$a ^ {1/3} + b ^ {1/3}$$ = ? => $$x + y =$$?

Let us attack statements

Statement 1:

$$a ^ {1/3} = b ^ {1/3} + 2$$ => $$x = y + 2$$ => $$x - y = 2$$

we have $$(x + y) (x - y) = 12$$
so $$x + y = 6$$ -> sufficient

Statement 2:
$$a = 64$$ => taking cube roots on both sides => $$a ^ {1/3}$$ = 4 => $$x = 4$$

using, $$x ^ 2 - y ^ 2 = 12$$ => $$4 ^ 2 - y ^ 2 = 12$$=> $$y ^ 2 = 4$$=>$$y = 2 or -2$$

so $$x + y = 4+2 = 6$$ or $$x - y = 4-2 = 2$$ => Not sufficient

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Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?  [#permalink]

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23 Nov 2018, 01:58
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Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?   [#permalink] 23 Nov 2018, 01:58
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