A square having sides parallel to the co-ordinate axes is inscribed in the region $$\{(x,y):x,y>0,y\leq -x^3+3x\}$$ if area of the square can be expressed as $$A^{\frac{1}{3}}+B^{\frac{1}{3}}$$ square units where $A$ and $B$ are integers and $A>B$ , then find $A$ and $B$
My Attempt
The vertices of the square can be assumed as $(x_{1},0)(x_{2},0),(x_{1},3x_{1}-x^3_{1}),(x_{2},3x_{2}-x^3_{2})$.
Now $3x_{1}-x^3_{1}=x_{2}-x_{1}=3x_{2}-x^3_{2}$
An attempt to solve leads to the equation $x^6_{1}-9x^4_{1}+21x^2_{1}-3=0$
After this i am not able to do.
As Dylan has pointed out, solving the equation gives $$x_1=\sqrt{3-2^{1/3}-2^{2/3}}$$ The height of the square is $$3x_1-x_1^3=3x_1-(3-2^{1/3}-2^{2/3})x_1=(2^{1/3}+2^{2/3})x_1$$ and the area is $$A=(2^{1/3}+2^{2/3})^2(3-2^{1/3}-2^{2/3})=(2^{2/3}+2^{4/3}+4)(3-2^{1/3}-2^{2/3})$$
Multiplying this out and simplifying gives $$A=6-3\cdot2^{2/3}=216^{1/3}-108^{1/3}$$