You need to write y in terms of x, hence, you should isolate `2y^3` to the left side, such that:

`-2y^3 = -x - 4`

Multiplying by -1 yields:

`2y^3 = x + 4`

Dividing by 2 yields:

`y^3 = (x+4)/2`

You need to take cube root both sides such that:

`root(3)(y^3) = root(3)((x+4)/2)`

`y = root(3)((x+4)/2)`

Hence, expressing y in terms of x yields `y = root(3)((x+4)/2).`

If you need to solve for x the equation `root(3)((x+4)/2) = 0` , you need to do the next steps such that:

`root(3)((x+4)/2) = 0 => (root(3)((x+4)/2))^3 = 0^3`

`(x+4)/2 = 0 => x + 4 = 0 => x = -4`

**Hence, if you need to express y in terms of x yields `y = root(3)((x+4)/2)` and if you need to solve for x the equation `y=0` yields `x = -4` .**

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