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# Does b^(1/3) = (2b - a)^(1/3)?

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Math Expert
Joined: 02 Sep 2009
Posts: 46305
Does b^(1/3) = (2b - a)^(1/3)? [#permalink]

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10 Sep 2017, 21:56
1
1
00:00

Difficulty:

45% (medium)

Question Stats:

53% (00:40) correct 47% (01:04) wrong based on 74 sessions

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Does $$\sqrt[3]{b} = \sqrt[3]{2b - a}$$?

(1) a > b
(2) a^2 > b

_________________
Intern
Joined: 27 Feb 2017
Posts: 3
Does b^(1/3) = (2b - a)^(1/3)? [#permalink]

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Updated on: 13 Sep 2017, 00:25
Bunuel wrote:
Does $$\sqrt[3]{b} = \sqrt[3]{2b - a}$$?

(1) a > b
(2) a^2 > b

For the above equation to hold, "b = 2b - a" must be true.
That is, -b = -a

If a > b, we know that the equation does't hold, it's enough.
If a^2 > b, we can't tell, since a,b may equal 2,3,4,5..., just not 1.

So, the answer is A (select to see).

Originally posted by orm1991 on 10 Sep 2017, 22:06.
Last edited by orm1991 on 13 Sep 2017, 00:25, edited 1 time in total.
Intern
Joined: 21 Nov 2016
Posts: 45
Re: Does b^(1/3) = (2b - a)^(1/3)? [#permalink]

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12 Sep 2017, 09:08
Bunuel wrote:
Does $$\sqrt[3]{b} = \sqrt[3]{2b - a}$$?

(1) a > b
(2) a^2 > b

This could be reduced to is b= 2b-a? or is b = a?

1. It clearly says a > b so b is not equal to a. Sufficient.
2. a^2 > b. Clearly insuff.

Re: Does b^(1/3) = (2b - a)^(1/3)?   [#permalink] 12 Sep 2017, 09:08
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