Area of square under a curve. 
A square having sides parallel to the coordinate axes is inscribed in the region.
  {$(x,y):x,y>0:y\le -x^3+3x$}.
  If the area of the square is written as $A^{1/3}+B^{1/3}$ square units where $A,B\in \Bbb Z$ and $A>B$, then find
  (i)$\sqrt{A-B}$
  (ii)Slope of line with x and y intercepts as $A,B$ respectively
  (iii)$A+B\over A-B$
  (iv)Circumradius of $\triangle OPQ$ where  $O$ is origin, $P(A^{1/3},0)$ and $Q(0,B^{1/3})$  

My approach:
Let the square be $KLMN$ where $K,L$ lie on x-axis with coordinates $(a,0);(b,0)$ respectively.
This will give coordinates of other points i.e. $M,N$ in terms of $a,b$.
Now taking out the length of all sides and equating them will give $a,b$.    
Problem with my approach:
I end up in a cumbersome equation involving $a,b$  
Tips: I know nothing more than simple differentiation and limits. So please avoid solutions with integration and other stuffs. 
PS: No need to downvote as I'm on the verge of getting banned. Please comment if you have any problem with the question.


Define $f(x):=3 x - x^3$. Obviously, the square must have a side length of $f(q)$, and we need to find $q$ that gives the largest square inside $f(x)$.
The line $y=f(q)$ intersects $y=f(x)$ at three points:
$$\left(q,f\left(q\right)\right)\\
\left(\frac{1}{2} \left(-\sqrt{3} \sqrt{4-q^2}-q\right),f\left(\frac{1}{2} \left(-\sqrt{3} \sqrt{4-q^2}-q\right)\right)\right)\\
\color{red}{\left(\frac{1}{2} \left(\sqrt{3} \sqrt{4-q^2}-q\right),f\left(\frac{1}{2} \left(\sqrt{3} \sqrt{4-q^2}-q\right)\right)\right)}\\
$$

We are only interested at the last point. The top left vertex of the square is $(q,f(q))$, and we need to ensure that $\sqrt{(f(q)-f(z))^2+(q-z)^2}=f(q)$, where $z=\frac{1}{2} \left(\sqrt{3} \sqrt{4-q^2}-q\right)$. So we need to find $q$ for the equation below:
  $$\sqrt{(f(q)-f(z))^2+(q-z)^2}-f(q)=0\tag{1}$$

You can use this implementation to get the intuition for the solution. 

This gives us two solutions:
$$\left\{\left\{q\to \sqrt{1+\sqrt[3]{2}}\right\},\left\{q\to \sqrt{3-\sqrt[3]{2}-2^{2/3}}\right\}\right\}$$
Which means the area of the maximum square is:
$$\bbox[5px,border:2px solid black]{f(q)^2 \iff f\left(\sqrt{3-\sqrt[3]{2}-2^{2/3}}\right)^2=6-3\times 2^{2/3}}$$

To get $A$ and $B$ set:
$$A^{1/3}=6\iff A=6^3=216\\
B^{1/3}=3\times2^{2/3} \iff B=(3\times2^{2/3})^3=108$$
Since $\sqrt[3]A+\sqrt[3]{-B}=\sqrt[3]A-\sqrt[3]{B}$, for reals, we have the solution:
$$\therefore{A=216\\B=-108}$$

The questions can now be easily answered:


*

*$\sqrt{216+108}=\sqrt{324}=18$

*Line passes through $(216,0)$ and $(0,-108)$, therefore: $y=\frac{-108}{216}(x-216)\iff y=\frac12x+108$

*$\frac{108}{324}=\frac13$

*The circumcircle is centered at:
$$\left(\frac62,-\frac{3\times2^{2/3}}{2}\right)$$ therefore the radius is $$r=\sqrt{3^2+\left(-\frac{3\times2^{2/3}}{2}\right)^2}=\sqrt{9+\frac{9}{2^{2/3}}}=\frac{3 \sqrt{1+2^{2/3}}}{\sqrt[3]{2}}$$

