Question 277573
Since {{{u=x^(1/3)^""}}}, we can convert to radical notation to get {{{u=root(3,x)}}}. It's just a different way to same thing.



Now because the possible solutions for 'u' (no pun intended) are 5 and -3, this means that {{{u=5}}} or {{{u=-3}}}



Let's find the solution(s) in terms of 'x' when {{{u=5}}}



{{{u=root(3,x)}}} Start with the given equation.



{{{5=root(3,x)}}} Plug in {{{u=5}}}



{{{(5)^3=(root(3,x))^3}}} Cube both sides to undo the cube root.



{{{(5)^3=x}}} Cube the cube root of 'x' to just get 'x'.



{{{125=x}}} Cube 5 to get 125.



So one solution in terms of 'x' is {{{x=125}}}


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Now let's find the solution(s) in terms of 'x' when {{{u=-3}}}



{{{u=root(3,x)}}} Start with the given equation.



{{{-3=root(3,x)}}} Plug in {{{u=-3}}}



{{{(-3)^3=(root(3,x))^3}}} Cube both sides to undo the cube root.



{{{(-3)^3=x}}} Cube the cube root of 'x' to just get 'x'.



{{{-27=x}}} Cube -3 to get -27.



So another solution in terms of 'x' is {{{x=-27}}}



So the solutions in terms of 'x' are {{{x=125}}} or {{{x=-27}}}