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 $753$

$r,s$ and $t$ are roots of the solution $8x^3+1001x+2008=0$

Sum of the roots $=-\left(\frac{coeffof{x}^2}{coeff.of{x}^3}\right)$

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

Product of the roots $=-\left(\frac{constantterm}{coeff.of{x}^3}\right)$

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

$r+s+t=0$

$r+s=-t$ --- ( 1 )

$s+t=-r$ --- ( 2 )

$r+t=-s$ --- ( 3 )

Now,

$(r+s{)}^3+(s+t{)}^3+(t+r{)}^3$ $=(-t{)}^3+(-r{)}^3+(-s{)}^3$ [ From ( 1 ), ( 2 ) and ( 3 ) ]

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

When $a+b+c=0$; then $a^3+b^3+c^3=3abc$

Since, $r+s+t=0;$ then $r^3+s^3+t^3=3rst$

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

$=-3\times (-251)$

$=753$