Question

If a + b + c = 15 and a2 + b2 + c2 = 83, then find the value of a3 +b3 + c3 - 3abc

Answer 1

a+ b+ c = 15
taking square of both side
a2 + b2 + c2 + 2(ab + bc + ca ) = 225
83 + 2 ( ab + bc + ca ) = 225
2 ( ab + bc + ca ) = 225- 83 = 142
ab +bc+ ca = 142÷2= 71
now ,
a3 + b3 + c3 - 3 abc
= ( a+ b + c) ( a2+ b2 + c2 - ab - bc - ca)
= 15× (83- 71 )
=15 ×12
= 180
hope this will help u ☺️

Answer 2

Answer:

a+ b+ c = 15, a2+b2+c2= 83

square both sides

a2 + b2 + c2 + 2(ab + bc + ca ) = 225

83 + 2 ( ab + bc + ca ) = 225

2 ( ab + bc + ca ) = 225- 83 = 142

ab +bc+ ca = 142÷2= 71

so now do,

a3 + b3 + c3 - 3 abc

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

= 15× (83- 71 )

=15 ×12

= 180