Question

What is the value of a³+b³+c³ −3abc ,if a + b + c = 12 and ab + bc + ca = 47

Answer 1

Given:

• a + b + c = 12
• ab + bc + ca = 47

To find:

⟶ a³ + b³ + c³ - 3abc

Formulas used:

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

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

Method:

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

Here we also need to find the value of a² + b² + c²

We know that;

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

Substituting the given values,

(12)² = a² + b² + c² + 2(47)

→ 144 = a² + b² + c² + 94

→ a² + b² + c² = 144 - 94

→ a² + b² + c² = 50

Now, we have got the value of a² + b² + c²

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

⇒ a³ + b³ + c³ - 3abc = 12(50 - (ab + bc + ca))

⇒ a³ + b³ + c³ - 3abc = 12(50 - 47)

⇒ a³ + b³ + c³ - 3abc = 12(3)

∴ a³ + b³ + c³ - 3abc = 36