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 :
Answer & explanation
Correct answer: option 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