It's better to generalize this

Consider the integral, $\displaystyle I = \int_0^\infty e^{-x^A} \text{ d}x$ . Substitute $\displaystyle x = u^{\frac{1}{A}} \Rightarrow \text{ d}x = \frac{1}{A} u^{\frac{1-A}{A}} \text{ d}u$ .
$\therefore I = \int_0^\infty e^{-u} \frac{1}{A} u^{\frac{1-A}{A}} \text{ d}u$ $\displaystyle I = \frac{1}{A} \int_0^\infty u^{\frac{1}{A} - 1} e^{-u} \text{ d}u$

$\displaystyle \Rightarrow I = \frac{1}{A} \Gamma \left( \frac{1}{A} \right) = \Gamma \left( \frac{1 + A}{A} \right)$
(Using the fact that $t \Gamma \left(t \right) = \Gamma \left(t + 1 \right)$ )

$\therefore \int_0^\infty e^{-x^A} \text{ d}x = \Gamma \left( \frac{1 + A}{A} \right)$ Put $A = 3$ to obtain the required answer as $\boxed{7}$ .