Relacionando $\sin{2\alpha}$ com $m$ têm-se: \begin{matrix} m^2 = (\sin{\alpha} + \cos{\alpha})^2 = \underbrace{\sin^2{\alpha} + \cos^2{\alpha}}_{1} + \underbrace{2\sin{\alpha} \cos{\alpha}}_{\sin{2\alpha}} &\therefore& \boxed{ \sin{2\alpha} = m^2 - 1} \end{matrix}Trabalhando agora o denominador, lembre-se: \begin{matrix}a^3 + b^3 = (a+b)(a^2 -ab + b^2) \end{matrix}Então, \begin{matrix} \sin^3{\alpha} + \cos^3{\alpha} = (\underbrace{\sin{\alpha} + \cos{\alpha}}_{m})(\underbrace{\sin^2{\alpha} - \sin{\alpha}\cos{\alpha} + \cos^2{\alpha}}_{1 - \large{\frac{m^2-1}{2}}}) = m\left(1 - \dfrac{m^2-1}{2} \right) \end{matrix}Substituindo na expressão do enunciado: \begin{matrix} y = \dfrac{m^2 - 1}{m\left(1 - \dfrac{m^2-1}{2} \right)} &\therefore& y = \dfrac{2(m^2 -1 )}{m(3 -m^2)} & \tiny{\blacksquare} \end{matrix}\begin{matrix}Letra \ (C) \end{matrix}