3.31: 

My equations are: 

\( a^4-4a \geq0 \)

\( x= (-a^2+ \sqrt{a^4-4a})/2 \)

But I don't know how to get the maximum value, I tried completing the square but it didn't work.

3.34:

I started with \( 20x=x^2+y^2 \),

then I got \( x^3-y^3-18xy = (x+y)(x^2-xy+y^2)-18y=(x+y)(20x-xy)-18xy=x((x+y)(20-y)-18y) \)

I got x=0 or y = 0, and (x+y)(20-y)-18xy=0. however, I can't figure out how to solve the equation.

3.35:

I substituted \( a= \sqrt[5]{x-6}\) and \(b=\sqrt[5]{39-x} \),

and I got ab(a+b)(a^2+b^2)=30. I tried c = ab and d=a+b, but It didn't really work out.