Question

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

• A. $751$
• B. $752$
• C. $753$
• D. $754$

Answer 1

The correct option is C $753$

We have,

$f\left(x\right)=8x^3+1001x+2008=0$

$f\left(x\right)=8x^3+0x^2+1001x+2008=0$

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

It is a cubic equation,

$Sumofroots=r+s+t=\frac{-\left(0\right)}{8}=0......\left(1\right)$

$Alsoproductofroots=r\times s\times t=rst=\frac{-\left(2008\right)}{8}=-251......\left(2\right)$

Now,

${(r+s)}^3+{(s+t)}^3+{(t+r)}^3={(-t)}^3+{(-r)}^3+{(-s)}^3=-(t^3+r^3+s^3)$

But we can get,

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

$\Rightarrow r^3+s^3+t^3=0+3(-251)=-753$

Now,

${(r+s)}^3+{(s+t)}^3+{(t+r)}^3=753$

Hence, this is the answer.