If a + b + c = 5, $a^2 + b^2+ c^2 = 27, $ and $a^3 + b^3 + c^3 = 125, $ then the value of $\frac{abc}{5}$ is :

-1

Given,

(a + b + c)² = 5²

a² + b² + c² + 2(ab + bc + ca) = 25

27 + 2(ab + bc + ca) = 25

2(ab + bc + ca) = -2

(ab + bc + ca) = -1

So,

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)]

125 – 3abc = 5 × [27 – ( -1 )]

125 – 3abc = 5 × 28

125 – 140 = 3abc

3abc = -15

abc = -5

So, $\frac{abc}{5}$ = $\frac{-5}{5}$ = -1