Let $r,s,t$ be the roots of equation $8x^3+1001x+2008=0.$ If value of $(r+s)^3+(s+t)^3+(t+r)^3$ is a three digit number with $a$, $b$ and $c$ as its digits, then value of $a+b+c$ is:

Since r, s, and t are roots of 8x^3 + 1001x + 2008 = 0, Vieta's formulas give r + s + t = 0, rs + st + rt = 1001/8, and rst = -2008/8. We need to evaluate the sum (r+s)^3 + (s+t)^3 + (t+r)^3, which simplifies to (-t)^3 + (-r)^3 + (-s)^3 = -(r^3 + s^3 + t^3). Using the identity r^3 + s^3 + t^3 = (r+s+t)^3 - 3(r+s+t)(rs+rt+st) + 3rst, we substitute the known values to get 0 - 0 + 3(-2008/8), which equals -753. Therefore, the required sum of cubes is -(-753) = 753. The sum of the digits a, b, and c of the number 753 is 7 + 5 + 3 = 15.

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