Question

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

Answer 1

To prove, a3+b3+c3−3abc=−25,
Given,
a+b+c=5, ab+bc+ca=10
∵(a+b+c)2=a2+b2+c2+2(ab+bc+ca)
∴(5)2=a2+b2+c2+2(10)
⇒25=a2+b2+c2+20
⇒a2+b2+c2=25–20
⇒a2+b2+c2=5
LHS=a3+b3+c3−3abc
=(a+b+c)(a2+b2+c2−ab−bc−ca)
=(5)[5−(ab+bc+ca)]
=5(5–10)=5(−5)=−25=RHS
Hence proved.