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# If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =

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Intern
Joined: 04 Jan 2017
Posts: 14
If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =  [#permalink]

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27 Jan 2017, 03:36
2
6
00:00

Difficulty:

65% (hard)

Question Stats:

61% (02:11) correct 39% (02:51) wrong based on 241 sessions

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If $$\frac{(x+1)^{1/3}-\frac{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0$$ , then x =

A. -3/2
B. -2/3
C. 0
D. 1/3
E. 1

Source: NOVA Math Review
Intern
Joined: 04 Jan 2017
Posts: 14
Re: If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =  [#permalink]

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27 Jan 2017, 03:40
I am a bit lost on this one. I keep getting -2/3 for an answer but OA is -3/2... My algebra skills are a bit rusty so I would be happy to see a step by step explanation of how to attack this beast...
Manager
Joined: 12 Mar 2013
Posts: 233
Re: If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =  [#permalink]

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27 Jan 2017, 03:44
3
1
denominator can't be zero. So (x+1)^1/3−1/3*x*(x+1)^−2/3 =0
(x+1)^1/3=1/3*x*(x+1)^−2/3
(x+1)^1/3= 1/3*x * 1/(x+1)^2/3
x+1= 1/3*x
3x+3=x
x=-3/2
A
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Intern
Joined: 04 Jan 2017
Posts: 14
Re: If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =  [#permalink]

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27 Jan 2017, 04:28
Thanks! I failed to realize that we can simply multiply away the denominator and got lost in my attempts to factor the expression out. That makes the whole expression much less daunting...
Manager
Joined: 12 Mar 2013
Posts: 233
Re: If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =  [#permalink]

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01 Feb 2017, 06:28
You are welcome brother.
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Re: If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =  [#permalink]

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04 Feb 2017, 17:25
1
denominator can't be zero. So (x+1)^1/3−1/3*x*(x+1)^−2/3 =0
(x+1)^1/3=1/3*x*(x+1)^−2/3
(x+1)^1/3= 1/3*x * 1/(x+1)^2/3
x+1= 1/3*x
3x+3=x
x=-3/2
A

I don't understand how you canceled out the 1/(x +1)^2/3
Math Expert
Joined: 02 Aug 2009
Posts: 7967
Re: If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =  [#permalink]

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04 Feb 2017, 19:10
Scyzo wrote:
If $$\frac{(x+1)^{1/3}-\frac{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0$$ , then x =

A. -3/2
B. -2/3
C. 0
D. 1/3
E. 1

Source: NOVA Math Review

Hi,

Denominator cannot be ZERO, otherwise the fraction will not be 0 but undefined..

So $$(x+1)^{1/3}-\frac{1}{3}x(x+1)^{-2/3}=0$$
Since exponent of (x+1) is -2/3, take it in denominator and change the sign..
OR multiply both sides by (x+1)^2/3.....
$$(x+1)^{1/3}*(x+1)^{2/3}-\frac{1}{3}x(x+1)^{-2/3}*(x+1)^{2/3}=0$$.....
$$(x+1)^{1/3+2/3}-\frac{1}{3}x(x+1)^{-2/3+2/3}=0$$..
$$(x+1)-\frac{1}{3}x=0$$....
$$x+1-\frac{x}{3}=0......\frac{2x}{3}=-1..... x=-1*\frac{3}{2}=\frac{-3}{2}$$
A
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Intern
Joined: 31 Jan 2017
Posts: 1
Re: If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =  [#permalink]

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10 Feb 2017, 14:11
How do you know when you can multiply the denominator by 0 (i.e. eliminate the denominator) and when you can't? If you have a fraction = zero, can you always just multiply the denominator by 0?
Intern
Joined: 07 Dec 2016
Posts: 6
Re: If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =  [#permalink]

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18 Jul 2018, 11:05
Scyzo wrote:
If $$\frac{(x+1)^{1/3}-\frac{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0$$ , then x =

A. -3/2
B. -2/3
C. 0
D. 1/3
E. 1

Source: NOVA Math Review

The hardest part is to imagine the 1/3 part. So better assign a value to it, let (x+1)^1/3 = a
Then the eqn eventually becomes

3a^3 -x = 0
now from here its easy to find the solution. 50 seconds to solve this qn!!! kudos
The trick lies in the imagination!!!
Intern
Joined: 16 May 2018
Posts: 2
Re: If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =  [#permalink]

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24 Jul 2018, 04:52
Take $$(x+1)^-2/3$$ common in the numerator:

Eqn then becomes :

$$(x+1)^-\frac{2}{3}$$[ (x+1) - x/3]/$$(x+1)^\frac{2}{3}$$

Cancelling $$(x+1)^\frac{-2}{3}$$, we have :

x+1-$$\frac{x}{3}$$ = 0
= 2x = -3
or x= $$\frac{-3}{2}$$
Re: If (x+1)^{1/3}-{1}{3}x(x+1)^{-2/3}}{(x+1)^{2/3}}=0 then x =   [#permalink] 24 Jul 2018, 04:52
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