Question

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

Answer 1

a+b+c=5-----(1)

ab+bc+ca=10-----(2)
square the (1)
(a+b+c)²=5²
a²+b²+c²+2ab+2bc+2ca=25
a²+b²+c²+2(ab+bc+ca) =25
a²+b²+c²+2*10=25 {from (2)}
a²+b²+c²=25-20
a²+b²+c²=5-----(3)
lhs =a³+b³+c³-3abc

= (a+b+c)(a²+b²+c²-ab-bc-ca)+3abc-3abc

=5*[5-(ab+bc+ca)] {from (1) and (3)}
=5*[5-10] { from (2)}
=5*(-5)
=-25
=rhs