If a + b + c = 9, and a2+b2+c2=35, find the value of a3+b3+c3−3abc.
a + b + c = 9
Squaring, we get
(a+b+c)2=(9)2⇒ a2+b2+c2+2(ab+bc+ca)=81⇒ 35+2(ab+bc+ca)=812(ab+bc+ca)=81−35=46∴ ab+bc+ca=462=23Now, a3+b3+c3−3abc=(a+b+c)[a2+b2+c2−(ab+bc+ca)]=9[35−23]=9×12=108