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Question

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

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Solution

To prove, a3+b3+c33abc=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=2520
a2+b2+c2=5
LHS=a3+b3+c33abc
=(a+b+c)(a2+b2+c2abbcca)
=(5)[5(ab+bc+ca)]
=5(510)=5(5)=25=RHS
Hence proved.

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