<s>[INST] Find "A" in this equation. $$ \sqrt[3] {A-15√3} + \sqrt[3] {A+15√3} = 4 $$
Find "A" ?
The way of exponentiation took too much time, is there any easier method?
 [/INST] Let $x=A-15\sqrt 3$ and $y=A+15\sqrt 3,$ then the equation becomes $$x^{1/3}+y^{1/3}=4.$$ Taking cubes of both sides gives $$x+y+3(xy)^{1/3}(x^{1/3}+y^{1/3})=4^3.$$ Now since $x+y=2A,$ and $x^{1/3}+y^{1/3}=4,$ the equation becomes $$2A+3(xy)^{1/3}(4)=4^3,$$ which gives $$6(xy)^{1/3}=32-A.$$ Now, cubing and substituting for $xy=A^2-15^2\cdot 3$ gives $$6^3(A^2-15^2\cdot 3)=(32-A)^3.$$ This is a cubic in $A,$ which in general is not so easy to deal with.
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