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

Question

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

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1 month 2021-08-18T10:04:05+00:00 2 Answers 0 views 0

1. ### Given :

• a + b + c = 15

• a² + b² + c² = 83

### ToFind:

• a³ + b³ + c³ – 3abc

### Solution:

We know that,

★ a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)

→ a³ + b³ + c³ – 3abc = (a + b + c) {(a² + b² + c²) – ab + bc + ca)} … i)

From the above identity, we’re required to find the values of a + b + c, + + and ab + bc + ca to get the value of + + 3abc

Now,

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

→ 15² = 83 + 2(ab + bc + ca)

→ 225 = 83 + 2(ab + bc + ca)

→ 225 – 83 = 2(ab + bc + ca)

→ 142 = 2(ab + bc + ca)

ab + bc + ca = 71

Put the value of ab + bc + ca in eq i)

★ a³ + b³ + c³ – 3abc = (a + b + c) {(a² + b² + c²) – ab + bc + ca)}

→ a³ + b³ + c³ – 3abc = 15 × (83 – 71)

→ a³ + b³ + c³ – 3abc = 15 × 12

a³ + b³ + c³ – 3abc = 180

The value of a³ + b³ + c³ – 3abc is 180.