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

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New post 27 Jan 2017, 03:36
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Question Stats:

60% (01:57) correct 40% (02:19) wrong based on 207 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
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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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New post 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...
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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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New post 27 Jan 2017, 03:44
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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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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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New post 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...
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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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New post 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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New post 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
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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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New post 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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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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New post 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?
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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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New post 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!!!
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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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New post 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 = &nbs [#permalink] 24 Jul 2018, 04:52
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