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Hi good people!,

 

Please help me out here...Solve for x in the equation:

 

\(x^{1 \over3}=4{1 \over x^3}\)

what I did was raise everything to another power of 3,

 

\(x=64{1 \over x^9}\),

 

then multiplied by \(x^9\)

 

\(x^{10}=64\),

 

then I'm stuck...

juriemagic  Apr 29, 2018
 #1
avatar+963 
+1

Here is my solution. 

 

\(x^\frac{1}{3} = y\)

 

y = 4 / y 

 

y^2 = 4

 

y = 2

 

\(x^\frac{1}{3}=2\)

x = the cube root of 2

 

I hope this helped,

 

Gavin

GYanggg  Apr 29, 2018
 #2
avatar+963 
+1

Never mind, I though the denominator was \(x^\frac{1}{3}\)

 

 

My answer is wrong. 

GYanggg  Apr 29, 2018
 #7
avatar+278 
+1

GYanggg,

 

thank you for trying...I do appreciate..

juriemagic  Apr 30, 2018
 #3
avatar
+1

Solve for x:
x^(1/3) = 4/x^3

Raise both sides to the power of three:
x = 64/x^9

Cross multiply:
x^10 = 64        Take the 10th root of both sides
x = 2^(3/5)      or        x= (-2)^(3/5)               

These 2 solutions are the only real solution to balance the equation.

Guest Apr 29, 2018
edited by Guest  Apr 29, 2018
 #4
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0

divide maybe?

Guest Apr 29, 2018
 #5
avatar+93866 
+2

x^{1 \over3}=4{1 \over x^3}

 

 

 

\(x^{1 \over3}=4{1 \over x^3}\\ x^{1 \over3}=4x^{-3}\\ x=4^3*x^{-9}\\ x^{10}=4^3\\ x=4^{3/10}\\ x=2^{3/5}\\ x=\sqrt[5]{8}\\ x\approx 1.5157 \)

 

this is the only positive real answer. 

I think our guest's negative answer also works.  It may not work with convention though

The cube root is in the question. so is a negative allowed.. I am not sure.

I think there are other complex answers as well.

Melody  Apr 30, 2018
edited by Melody  Apr 30, 2018
edited by Melody  Apr 30, 2018
 #8
avatar+278 
0

Hi Melody,

 

a Million thank you's...I think your answer is closest to what I believe the answer should be. I do appreciate!

juriemagic  Apr 30, 2018
 #6
avatar+7517 
0

One more try:

a\(x^{\frac{1}{3}}=4{\frac{1}{x^3}}\)

 

definition:

\({\color{black}is}\ 4\frac{1}{3}=4 +\frac{1}{3}\ {\color{black}so\ is}\ 4\frac{1}{x^3}=4 +\frac{1}{x^3}\)

 

\(x^{\frac{1}{3}}=4+{\frac{1}{x^3}}\)         |  \(\times x^3\)

\(x^{\frac{10}{3}}=4x^3+1\)

\(f(x)=4x^3-x^{\frac{10}{3}}+1 \neq 0\)

 

The function has a minimum in P (0;1).
The function has no zero.

laugh  !

asinus  Apr 30, 2018
edited by asinus  Apr 30, 2018
 #9
avatar+278 
+1

asinus,

 

wow, this is something else!!..would never have thought of going that route. I do appreciate your help! Thank you..

juriemagic  Apr 30, 2018
 #12
avatar
0

If you smoke some whackyweed you might think of going that route. 

Asinus is a natural space cadet, so he probably doesn’t need to smoke any.

Maybe it would help, if he did.

Guest May 1, 2018

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