If a+b+c= 5 and ab+bc+ca =10,

To prove, $a^{3}+b^{3}+c^{3}-3 a b c=-25$

Given, $a+b+c=5, a b+b c+c a=10$

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

$\therefore$ $(5)^{2}=a^{2}+b^{2}+c^{2}+2(10)$

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

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

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

$\mathrm{LHS}=\mathrm{a}^{3}+b^{3}+c^{3}-3 a b c$

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

$=(5)[5-(a b+b c+c a)]$

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

Hence proved.