Find the value of $?$ in the following expression: $\frac{(0.73)^3 + (0.27)^3}{(0.73)^2 - 0.73 \times 0.27 + (0.27)^2} \times (\frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56})^{-1} = ?$

1) The algebraic identity $\frac{a^3 + b^3}{a^2 - ab + b^2} = a + b$. Here $a=0.73, b=0.27$, so $a+b = 1.0$. 2) Telescoping series: $\frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} = (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + \dots + (\frac{1}{7} - \frac{1}{8}) = \frac{1}{3} - \frac{1}{8} = \frac{5}{24}$. 3) Taking the inverse: $(\frac{5}{24})^{-1} = \frac{24}{5} = 4.8$. 4) Result: $1.0 \times 4.8 = 4.8$.