A
pity to leave this one without a derivation of armando’s findings.
Rather than invoke giants’ shoulders etc. here’s a school maths
approach which starts from armando’s first solution.

Since C is not zero, let x = A/C and y = B/C then A^{3} + B^{3}
= 31C^{3}

transforms to the cubic curvex^{3}
+ y^{3} = 31with a rational
point

P(x_{1}, y_{1}) = P(137/42, -65/42) corresponding to armando’s
result.

We need only find the point where the tangent at P meets the

curve again. It will also be rational and give us a second solution.

Gradient: dy/dx = -x^{2}/y^{2}, so the tangent at P(x_{1},
y_{1}) will have the

equation:y = mx + c wherem = -(137/-65)^{2} = 18769/4225

and c = -65/42 - m(137/42) = 54864/4225

Where it crosses the curve again,x^{3}
+ (mx + c)^{3} = 31, which

We know there is a double root at P, (x = x_{1}), so the LHS factors: (m^{3} + 1)x^{3} +3m^{2}cx^{2}
+ 3 mc^{2}x + c^{3} – 31 = (m^{3} + 1)(x – x_{1})^{2}(x
– x_{2})

and by equating the constant terms, (or other coefficients), the

remaining root, x_{2}, can be found:c^{3} – 31 = (m^{3} + 1)(-
x_{1}^{2}x_{2})

givingx_{2} = (31 – c^{3})/(x_{1}^{2}(m^{3}
+ 1))

Substituting for m, c, and x_{1} : x_{2}
= 277028111/119531076.

Thusy_{2} = 316425265/119531076 and these
give another trio:

(A, B, C) = (277028111, 316425265, 119531076) as armando found.

In the same way, this new solution spawns other solutions, but will

they have A, B and C all positive and is there a smaller solution?

It’s worth drawing the cubic curve x^{3} + y^{3} = 31 and a few
tangents. Its

shape provides some useful insights and a lot of fun.

(‘Rational Points on Elliptic Curves’ by Silverman & Tate was very useful – many years ago).