Question

Simplify: ((2x+1)^(2/3)-((3(x^-1))/((2x+1)^(1/3))/((2x+1)^2)

Answer 1

According to Mathematica v8, the two open parentheses in red are not balanced.

Answer 2

((2x+1)^(2/3)-((3x^(-1))/(2x+1)^(1/3)))/((2x+1)^(2)) I think that's the one that's correct. I'll try to make an equation easier to read.

Answer 3

It won't let me put the exponents as fractions.

Answer 4

just draw it

Answer 5

((2x+1)^(2/3)-((3x^(-1))/(2x+1)^(1/3)))/((2x+1)^(2))\[\left((1+2 x)^{2/3}-\left((3x{}^{\wedge}(-1))\left/(1+2 x)^{1/3}\right.\right)\right)/(1+2 x)^2 \]\[\left((1+2 x)^{2/3}-\left(\frac{3}{x}/(1+2 x)^{1/3}\right)\right)/(1+2 x)^2 \]\[\left((1+2 x)^{2/3}-\frac{3}{x (1+2 x)^{1/3}}\right)/(1+2 x)^2 \]\[\left(\frac{-3+x+2 x^2}{x (1+2 x)^{1/3}}\right)/(1+2 x)^2 \]\[\frac{-3+x+2 x^2}{x (1+2 x)^{7/3}} \]\[\frac{2 x^2+x-3}{x (2 x+1)^{7/3}} \]

Answer 6

Thanks!

Answer 7

Thank you for the medal.