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

(N/A) We know the algebraic identity:
$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$= (a+b+c)[a^2+b^2+c^2-(ab+bc+ca)]$
Given $a+b+c=5$ and $ab+bc+ca=10$,we substitute these values:
$= 5[a^2+b^2+c^2-10]$
Now,we find $a^2+b^2+c^2$ using the identity $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$:
$(5)^2 = a^2+b^2+c^2+2(10)$
$25 = a^2+b^2+c^2+20$
$a^2+b^2+c^2 = 25-20 = 5$
Substituting this value back into the expression:
$a^3+b^3+c^3-3abc = 5(5-10) = 5(-5) = -25$.
Hence,it is proved.