Question

Find the value of a cube + b cube + c cube -3abc of a+b+c=5 and a sq+ bad+c sq=29

Answer 1

Answer:

155

Step-by-step explanation:

Given that a + b + c = 5 and a² + b² + c²= 29

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

5²=29 + 2(ab + bc + ca)

25 - 29 = 2(ab + bc + ca)

-4/2 = (ab + bc + ca)

(ab + bc + ca) = -2

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 = 5(29 - (-2))

a³ + b³ + c³ - 3abc = 5 × 31 = 155