Question

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

Answer 1

Answer:

hey mate here is ur answer

Step-by-step explanation:

a³ + b³ + c³ - 3abc = ( a + b + c ) ( a² + b² + c² - ( ab + bc + ca ). = ( 15 ) ( 83 - ( 71 ) ). = 15 × 12. = 180

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Answer 2

Given :

• a + b + c = 15

• a² + b² + c² = 83

To Find :

• 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, a² + b² + c² and ab + bc + ca to get the value of a³ + b³ + c³ - 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

$\therefore$ The value of a³ + b³ + c³ - 3abc is 180.