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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**+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**+1 **

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

My answer is wrong.

GYanggg Apr 29, 2018

#3**+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

#5**+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

#8**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**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.

!

asinus Apr 30, 2018

#9**+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