Question

Let $k$ are roots of equation $8x^{3}+1001x+2008=0.$ Then value of $\left(r + s\right)^{3}+\left(s + t\right)^{3}+\left(t + r\right)^{3}$ is $7k3$ (where k is at ten's place). Then $k=$

Answer 1

$r+s+t=0, \, rst=\frac{- 2008}{8}=-251$
Now Let
$r+s=x$
$s+t=y$
$t+r=z$
$\therefore \, \, x+y+z=2\left(r + s + t\right)=0$
$\therefore \, \, \, x^{3}+y^{3}+z^{3}=3xyz$
$=3\left(r + s\right) \, \left(s + t\right) \, \left(t + r\right)$
$=3\left(- t\right) \, \left(- r\right) \, \left(- s\right)=-3 \, rst$
$=-3\times \left(- 251\right)=753$