Yes, I'll solve it now, I just wanted to make sure I had the right question.
Anyways, we have: \(4x^{1 \over 3} - 2{x \over x^{2\over 3}} = 7 + \sqrt {3x}\)
First, note that \({x \over {x ^ {2 \over 3}}} = {x^{3 \over 3} \over x^{2 \over 3}} = x^{1 \over 3}\)
This means we have: \(4x^{1 \over 3} - 2{x^{1\over 3}} = 7 + \sqrt {3x}\).
Subtracting like terms gives us: \(2x^{1 \over 3} = 7 + \sqrt{3x}\)
To solve, let \(y = x^{1 \over 3}\). This gives \(2y = 7 + \sqrt{3y^3}\)
We can convert this to \(0=-3y^3+ 4y^2-28y+49\).
Solving for y gives us \(y \approx 1.65543\), meaning \(x = \sqrt[3]y \approx1.18296\).
However, plugging this in doesn't work, meaning the solution is imaginary....
According to WA, its \(\approx -0.161 + 3.137i\).
However, you should probably plug it in the original equation to see if it works.....