If 4 [ ( 17 )^3 - ( 7 )^3 ] ( 17^2 + 7^2 + p ) = 40, then what is the value of p?

Answer: Option C

$$\begin{array}{rl}& \frac{4\left[{\left(17\right)}^3-{\left(7\right)}^3\right]}{\left({17}^2+7^2+p\right)}=40\\ & \text{We know,}\\ & \left[a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right]\\ & \frac{4\left[\left(17-7\right)\left({17}^2+7^2+17\times 7\right)\right]}{{17}^2+7^2+p}=40\\ & 17^2+7^2+119=17^2+7^2+p\\ & p=119\end{array}$$