Question

How do you solve `x(x^2-3x)^(1/3) + 2(x^2-3x)^(4/3) = 0`?

Answer 1

Temporarily replace `(x^2-3x)` with `k`

`x(x^2-3x)^(1/3)+2(x^2-3x)^(4/3) = 0`

can be re-written (using the temporary substitution) as
`x*k^(1/3) + 2k*k^(1/3) = 0`

`(k^(1/3))*(x+ 2(x^2-3x)) = 0`

`(k^(1/3))*(2x^2 - 5x) = 0`

One of the terms
`k^(1/3) = root(3)(x^2-3x)`
or
`2x^2-5x`
must equal `0`

`root(3)(x^2-3x)`
can only equal `0`
if `x^2-3x = 0`
`rarr x = 0`
or
`x=3`

`2x^2-5x`
can only equal `0`
if `x=0`
or
`x = 5/2`

Therefore `x epsilon { 0, 2 1/2, 3}`