If a + b + c = 6, a² + b² + c² = 32, and a³ + b³ + c³ = 189, then the value of 4abc is:

12

a + b + c = 6

a² + b² + c² = 32

a³ + b³ + c³ = 189

= (a + b + c)² - 2(ab + bc + ca) = 32

= 6² - 2(ab + bc + ca) = 32

= -2(ab + bc + ca) = 32 - 36

= -2(ab + bc + ca) = -4

= (ab + bc + ca) = 2      ....(a)

a³ + b³ + c³ = 189

= (a + b + c) (a² + b² + c² - (ab + bc + ca)) + 3abc = 189

= 6 × (32 - 2) + 3abc = 189 (From a)

= 6 × 30 + 3abc = 189

= 3abc = 189 - 180

= 4abc = \(\frac{9 × 4}{ 3}\)

= 4abc = 12