I noticed that maybe you shouldn't think of those cube roots as products of root 2 and 3. What I started with was cbrt(1*1) + cbrt(1*2) + cbrt(2*2). It goes until cbrt(3*3) + cbrt(3*4) + cbrt(4*4). I'll come back when I find something.

I reduced all the cube roots to be multiples of cr(2) and cr(3).

For example cr(12) = cr(2)*cr(2)*cr(3).

I let x = cr(2) and y = cr(3) so cr(12) = x^2y.

Now I had an equation where the terms in the denominator are powers of x and y. At this point I tried to find some clever factoring strategy, but alas I couldn't. Left with only getting a common denominator and expanding out 27 terms in the numerator, I let my impatience get the better of me and decided to call a halt. If anyone has a better approach, I like to see it.

Excellent!

a^3 - b^3 = (a -b)(a^2 + ab + b^2) was lurking there all the time and I didn't see how it fit - but you did. Good work.

Ian

I noticed that maybe you shouldn't think of those cube roots as products of root 2 and 3. What I started with was cbrt(1*1) + cbrt(1*2) + cbrt(2*2). It goes until cbrt(3*3) + cbrt(3*4) + cbrt(4*4). I'll come back when I find something.

I reduced all the cube roots to be multiples of cr(2) and cr(3).

For example cr(12) = cr(2)*cr(2)*cr(3).

I let x = cr(2) and y = cr(3) so cr(12) = x^2y.

Now I had an equation where the terms in the denominator are powers of x and y. At this point I tried to find some clever factoring strategy, but alas I couldn't. Left with only getting a common denominator and expanding out 27 terms in the numerator, I let my impatience get the better of me and decided to call a halt. If anyone has a better approach, I like to see it.