Question

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:

• A. $15$
• B. $27$
• C. $21$
• D. $18$

Answer 1

The correct option is A $15$

$r+s+t=0$; Since co-efficient of $x^2$ is $0$

$rst=\frac{-2008}{8}=-251$

Now, let $r+s=l,s+t=m,t+r=n,$

$\Rightarrow l+m+n=2(r+s+t)=0$

$l^3+m^3+n^3=3lmn$

$\therefore (r+s{)}^3+(s+t{)}^3+(t+r{)}^3=3(r+s\left)\right(s+t\left)\right(t+r)$

$=-3rst=3\times 251=753$

Hence, value of $a+b+c$ is $15$