If $a + b + c = 5$ and $ab + bc + ca = 10$, then prove that $a^3 + b^3 + c^3 - 3abc = -25$.

Identity: $a^{3} + b^{3} + c^{3} - 3abc$

$$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$

$$\text{Also, } (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$

$$25 = a^2 + b^2 + c^2 + 20$$

$$\Rightarrow a^2 + b^2 + c^2 = 5$$

$$\text{So, } a^2 + b^2 + c^2 - ab - bc - ca = 5 - 10 = -5$$

$$\text{Therefore, } a^3 + b^3 + c^3 - 3abc = 5 \times (-5) = -25.$$

Concept Applied
$a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$; when $a + b + c = 0$,
 the expression equals $3abc$.