If a + b + c = 7, a^2 + b^2 + c^2= 35 and a^3 + b^3 + C^3 = 151 then find the value of abc?

Question

If a, b, c are three real numbers such that a + b + c = 7, a^2 + b^2 + c^2= 35 and a^3 + b^3 + C^3 = 151. Find the value of abc?

Answer 1

(a+b+c)² =49
Again, (a+b+c)² = a² + b² + c² + 2(ab + bc + ac)
Thus, 35 + 2(ab + bc + ac) = 49
Or, (ab + bc + ac) = 7
Again, we know, a³ + b³ + c³ - 3abc = (a+b+c){(a² + b² + c²) - (ab + bc + ac)}
Or, 151 - 3abc = 7(35 - 7)
Or, -3abc = 7×28
Or, abc = 196/(-3)
Or, abc = -15

Hence, Answer is (-15)