Question

Find the value of a3 + b3 +c3

– 3abc, when a + b + c = 8 and a b + b c +c a = 25
plz answer properly
and step by step plzzzzz​

Answer 1

Answer:

-88

Step-by-step explanation:

Given, a + b + c = 8

On Squaring both sides, we get

=> (a + b + c)² = (8)²

=> a² + b² + c² + 2ab + 2bc + 2ca = 64

=> a² + b² + c² + 2(ab + bc + ca) = 64

=> a² + b² + c² + 2(25) = 64

=> a² + b² + c² + 50 = 64

=> a² + b² + c² = 64 - 50

=> a² + b² + c² = 14

Now,

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)]

=> a³ + b³ + c³ - 3abc = (8) [14 - 25]

=> a³ + b³ + c³ - 3abc = 8 (-11)

a³ + b³ + c³ - 3abc = - 88

Hope it helps!

Answer 2

Answer:

thank you for your answer