If $a+b+c=9$ and $ab+bc+ca=26$, then $a^3+b^3+c^3-3abc$ equals:

First $a^2+b^2+c^2=81-52=29$. The identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ then gives $9(29-26)=27$. (Omitting the subtraction gives $9\times 29=261$, and using $(a+b+c{)}^2$ in place of $a^2+b^2+c^2$ gives $9\times 55=495$.) With a, $b,c=2,3,4:8+27+64-3(24)=99-72=27$.