Question

a+b+c=6 and ab+bc+ca=11 then a^3+b^3+c^3-3abc is equal to?

Answer 1

Answer:

18

Step-by-step explanation:

Given :

a + b + c = 6 and ab + bc + ca = 11,

then a³ + b³ +c³ - 3abc is equal to??

Solution :

We know that,

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

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

⇒ (6)² = a² + b² + c² + 2(ab + bc + ac)

⇒ 36 = a² + b² + c² + 2(11)

⇒ 36 = a² + b² + c² + 22

⇒ 36 - 22 = a² + b² + c²

⇒ 14 = a² + b² + c² ...(i)

__

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

⇒ (6) [14 - 11]

⇒ 6 × 3 = 18