Answer:
x=6
Step-by-step explanation:
[tex]\sqrt[3]{(x+2)^{2} } =\sqrt[3]{(x+58)} \\\\(\sqrt[3]{(x+2)^{2} })^{3} =(\sqrt[3]{(x+58)} )^{3} \\\\(x+2)^{2}=(x+58)\\x^{2} +4x+4=x+58\\\\x^{2} +3x-54=0[/tex]
Solving this quadratic equation:
x'=6
x''=-9
But x>0, then the solution is x=6
S={6}
