Question

A+B+C=5 AB+BC+CA=16 PROVE A^3+B^3+C^3-3ABC=25

Answer 1

Correct Question :-

If a + b + c = 5, ab + bc + ca = 16. Prove a³ + b³ - 3abc = -115

Solution :-

a + b + c = 5

ab + bc + ca = 16

First find the value of a² + b² + c²

We know that

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

Here

• a + b + c = 5

• ab + bc + ca = 16

By substituting the values in the udentity

(5)² = a² + b² + c² + 2(16)

25 = a² + b² + c² + 32

25 - 32 = a² + b² + c²

- 7 = a² + b² + c²

a² + b² + c² = - 7

We know that

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

Here

• a + b + c = 5

• a² + b² + c² = - 7

• ab + bc + ca = 16

By substituting the values

a³ + b³ + c³ - 3abc = (5){- 7 - (16)}

a³ + b³ + c³ - 3abc = 5(- 7 - 16)

a³ + b³ + c³ - 3abc = 5(-23)

a³ + b³ + c³ - 3abc = - 115

Hence proved

Answer 2

Questions—

a + b + c = 5, ab + bc + ca = 16. Prove a³ + b³ - 3abc = -115

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Given—

✴ a + b + c = 5

✴ ab + bc + ca = 16

We know,

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

Finding

a² + b² + c² to put in above Formula

A/Q

✴ a + b + c = 5

✴ ab + bc + ca = 16

Substitue the value

=> a² + b² + c² + 2(16) = (5)²

=>a² + b² + c² + 32 = 25

=> a² + b² + c² = 25 - 32

=>a² + b² + c² = - 7

We know,

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

✴ a + b + c = 5

✴ a² + b² + c² = - 7

✴ ab + bc + ca = 16

Let substitute the values—

a³ + b³ + c³ - 3abc = (5){- 7 - (16)}

a³ + b³ + c³ - 3abc = 5(- 7 - 16)

a³ + b³ + c³ - 3abc = 5(-23)

a³ + b³ + c³ - 3abc = - 115

Therefore. It is proved

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