If $a+b+c=5$ and $a b+b c+c a=10,$ then prove that $a^{3}+b^{3}+c^{3}-3 a b c=-25.$

We know that,

$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$=(a+b+c)\left[a^{2}+b^{2}+c^{2}-(a b+b c+c a)\right]$

$=5\left\{a^{2}+b^{2}+c^{2}-(a b+b c+c a)\right\}$

$=5\left(a^{2}+b^{2}+c^{2}-10\right)$

Now, $\quad a+b+c=5$

Squaring both sides, we get

$(a+b+c)^{2}=5^{2}$

$\Rightarrow a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=25$

$\therefore \quad a^{2}+b^{2}+c^{2}+2(10)=25$

$\Rightarrow \quad a^{2}+b^{2}+c^{2}=25-20=5$

Now, $a^{3}+b^{3}+c^{3}-3 a b c=5\left(a^{2}+b^{2}+c^{2}-10\right)$

$=5(5-10)=5(-5)=-25$

Hence, proved.