Let r,s, and t be root …

Let {tex}f(x)=8x^{ 3 }+1001x+2008=0\Rightarrow f(x)=8x^{ 3 }+0{ x }^{ 2 }+1001x+2008=0{/tex}

Given the roots of this equation are r,s,t.

Now using the relationship between the zeroes and coefficients of a cubic polynomial we have, 

 sum of the roots= {tex}r+s+t=\frac { -\left( 0 \right) }{ 8 } =0{/tex}.........................(i)

Also product of the roots={tex}rst=\frac { -\left( 2008 \right) }{ 8 } =-251{/tex}............(ii)

Now {tex}(r+s)^{ 3 }+(s+t)^{ 3 }+(t+r)^{ 3 }=(-t)^{ 3 }+(-r)^{ 3 }+(-s)^{ 3 }=-\left( t^{ 3 }+r^{ 3 }+s^{ 3 } \right) {/tex}.............(iii)          [using (i)]

But we can get  {tex}r^{ 3 }+s^{ 3 }+t^{ 3 }=\left( r+s+t \right) \left( r^{ 2 }+s^{ 2 }+t^{ 2 }-rs-st-tr \right) +3rst{/tex}

{tex}\Rightarrow r^{ 3 }+s^{ 3 }+t^{ 3 }=0+3\left( -251 \right) =-753{/tex}   [using (i) and (ii)]

Therefore from (iii) we have {tex}(r+s)^{ 3 }+(s+t)^{ 3 }+(t+r)^{ 3 }=-\left( t^{ 3 }+r^{ 3 }+s^{ 3 } \right) =-\left( -753 \right) =753{/tex}