Sum of Their Cubes

We could work out the actual values of $a$ and $b$ , but it's neater to do the following.

Factorising, we have $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)((a+b)^2-3ab)$

Now we just substitute $a+b=10$ and $ab=5$ to get $a^3+b^3=10\times (10^2-3 \times 5)=\boxed{850}$ .

$\begin{aligned} (a+b)^3 & = a^3 + 3a^2b + 3ab^2 + b^3 \\ \implies a^3+b^3 & = (a+b)^3 - 3ab(a+b) \\ & = 10^3 - 3\times 5 \times 10 \\ & = \boxed{850} \end{aligned}$