Nice

The expression $c^{3} = a^{3} + b^{3}$ can be rewritten as: $c^{3} = (a + b)(a^{2} - ab + b^{2}) = (a + b)(c^{2} + 2ab \cdot cos(C) - ab)$ (i). By the Triangle Inequality, we must satisfy $a + b > c$. Thus, let $a + b = Kc$ (for $K > 1$) and substitute this value into (i) to obtain: $c^{3} = Kc(c^{2} + 2ab \cdot cos(C) - ab)$; or $c^{2} = K(c^{2} + 2ab \cdot cos(C) - ab)$; or $(1/K - 1)c^{2} + ab = 2ab \cdot cos(C)$; or $cos(C) = \frac{(1/K - 1)c^{2} + ab}{2ab}$ (ii). What follows is a proof by contradiction. Suppose $0 < C \le 60^{\circ}$, which in turn yields $1/2 \le cos(C) < 1$. Substituting (ii) into this inequality produces: $1/2 \le [(1/K - 1)c^{2} + ab] / 2ab < 1$; or $ab \le (1/K - 1)c^{2} + ab < 2ab$; or $0 \le (1/K - 1)c^{2} < ab$ (iii). Since the quantity 1$/K - 1$ is strictly negative, (iii) is only satisfied when $c = 0$ (a contradiction). Hence, $60^{\circ} < C < 180^{\circ}$. QED