Question

Let $r,s,t$ be roots of the equation $8x^3+1001x+2008=0$. The value of $(r+s{)}^3+(s+t{)}^3+(t+r{)}^3$is

• A. $251$
• B. $751$
• C. $735$
• D. $753$

Answer 1

The correct option is D $753$

The given equation can be written as $8x^3+0x^2+1001x+2008=0$

The sum of the roots is,

$r+s+t=\frac{-0}{8}$

$=0$

The product of the roots is,

$rst=-\frac{2008}{8}$

$=-251$

Now, ${\left(r+s\right)}^3+{\left(s+t\right)}^3+{\left(t+r\right)}^3={\left(-t\right)}^3+{\left(-r\right)}^3+{\left(-s\right)}^3$

$=-\left(t^3+r^3+s^3\right)$

But $r^3+s^3+t^3=\left(r+s+t\right)\left(r^2+s^2+t^2+-rs-st-tr\right)+3rst$

$=0+3\left(-251\right)$

$=-753$

So, $-\left(t^3+r^3+s^3\right)=-\left(-753\right)$

$=753$