If (a - b) = 3, (b - c) = 5 and (c - a) = 1, then the value of $$\frac{a^3 + b^3 + c^3 - 3abc}{a + b + c}$$ is

it is given that (a - b) = 3, (b - c) = 5 and (c - a) = 1 

we need to find the value of $$\frac{a^3 + b^3 + c^3 - 3abc}{a + b + c}$$ 

as we know that $${a^3 + b^3 + c^3}$$ = $$(a+b+c)({a^2 + b^2 + c^2 - ab - bc - ac})$$ ........(5)

and $${(a-b)^2 = a^2 + b^2 - 2ab}$$.......(1)

$${(b-c)^2 = b^2 + c^2 - 2cb}$$...........(2)

$${(c-a)^2 = a^2 + c^2 - 2ac}$$..........(3)

adding 1 , 2 and 3 

17.5 =  $$({a^2 + b^2 + c^2 - ab - bc - ac}$$ .....(4)

Now using 4 and 5 statement 

$$\frac{a^3 + b^3 + c^3 - 3abc}{a + b + c}$$  = 17.5