Question

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

Answer 1

Solution :

We know that

a³ + b³ + c³ - 3abc

= (a + b + c)[(a² + b² + c²) - (ab + bc + ca)]...(I)

Now,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

15² = 83 + 2(ab + bc + ca)

142 = 2(ab + bc + ca)

ab+ bc + ca

= 142/2

= 71

Substituting the value of ab + bc + ca in (i),

we get

a³ + b³ + c³ - 3abc

= 15 (83 - 71)

= 15 x 12

= 180.