Simplifying, logs, and solving

Please show work! I need to learn how to do these type of problems by looking at the work.

Simplify: { (x^(7/3) + x^(4/3)) / (x^2-2x+3)} / (1/x^3)

Solve: (x+1)^3=(x-1)^3+26

Solve: x^(2/5) + 5 x^(1/5) + 6=0

log x^(1/3)= (log x)^1/2

((x^2-1) / x)^2 - ((x^2-1) / x)-2=0

Thanks! Best answer goes to the person who shows work and answers all questions!

The first problem is not an equation, so it can't be solved.

(x + 1)^3 = (x - 1)^3 + 26
x^3 + 3x^2 + 3x + 1 = x^3 - 3x^2 + 3x - 1 + 26
6x^2 = 24
x^2 = 4
x = ± 2

x^(2/5) + 5x^(1/5) + 6 = 0
(x^(1/5) + 2)(x^(1/5) + 3) = 0
x^(1/5) + 2 = 0 or x^(1/5) + 3 = 0
x^(1/5) = -2..........x^(1/5) = -3
x = -32 or x = -243

log x^(1/3) = (log x)^(1/2)
(1/3) log x = (log x)^(1/2)
(1/9) (log x)^2 = log x
(1/9) (log x)^2 - log x = 0
log x ((1/9) log x - 1) = 0
log x = 0 or (1/9) log x - 1 = 0
x = 1.............(1/9) log x = 1
............................log x = 9
............................x = 10^9

((x^2 - 1)/x)^2 - ((x^2 - 1)/x) - 2 = 0
((x^2 - 1)/x - 2)((x^2 - 1)/x + 1) = 0
(x^2 - 1)/x = 2.......(x^2 - 1)/x = -1
x^2 - 1 = 2x............x^2 - 1 = -x
x^2 - 2x - 1 = 0..........x^2 + x - 1 = 0
x^2 - 2x = 1..............x ≈ 0.618 or -1.618
x^2 - 2x + 1 = 1 + 1
(x - 1)^2 = 2
x - 1 = ± √2
x = 1 ± √2