Question

if a+b+c=6 and ab+bc+ca=11 then find value of a^3+b^3+c^3-3abc

Answer 1

a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
(a+b+c)^2 =( a^2+b^2+c^2)+2(ab+bc+ca)
(6)^2= ( a^2+b^2+c^2)+2(11)
36= ( a^2+b^2+c^2)+22
36-22= ( a^2+b^2+c^2)
14= ( a^2+b^2+c^2)
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))
=(6)(14-(11))
=(6)(3)
=18
this took long to type but it's easy to write
I should have just taken a pic of the working