Question

Let $\alpha ,\beta ,\gamma$ be the roots of the equation $8x^3+1001+x+2008=0$. Then the value of $(\alpha +{\beta }^2)+(\beta +{\gamma }^3)+(\gamma +{\alpha }^3)$, is:

• A. $251$
• B. $753$
• C. $735$
• D. $253$

Answer 1

The correct option is B $753$

$\alpha ,\beta ,\gamma$ are the roots of the equation $8x^3+1001x+2008=0$

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

$\Rightarrow$ $\alpha +\beta +\gamma =\frac{0}{8}=0$

$\Rightarrow$ $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{coeff.ofx}{coeff.x^3}=\frac{1001}{8}$

$\Rightarrow$ $\alpha \beta \gamma =-\frac{constantterm}{coeff.of{x}^3}=-\frac{2008}{8}=-251$

$\alpha +\beta +\gamma =0$ then,

$\alpha +\beta =-\gamma$

$\beta +\gamma =-\alpha$

$\gamma +\alpha =-\beta$

Now,

$(\alpha +\beta {)}^3+(\beta +\gamma {)}^3+(\gamma +\alpha {)}^3$

$\Rightarrow$ $(\alpha +\beta {)}^3+(\beta +\gamma {)}^3+(\gamma +\alpha {)}^3$ $=(-\gamma {)}^3+(-\alpha {)}^3+(-\beta {)}^3$

$\Rightarrow$ $(\alpha +\beta {)}^3+(\beta +\gamma {)}^3+(\gamma +\alpha {)}^3$$=-({\alpha }^3+{\beta }^3+{\gamma }^3)$

We know, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\Rightarrow$ $(\alpha +\beta {)}^3+(\beta +\gamma {)}^3+(\gamma +\alpha {)}^3$ $=-[3\alpha \beta \gamma +(\alpha +\beta +\gamma \left)\right({\alpha }^2+{\beta }^2+{\gamma }^2-\alpha \beta -\beta \gamma -\gamma \alpha \left)\right]$

$\Rightarrow$ $(\alpha +\beta {)}^3+(\beta +\gamma {)}^3+(\gamma +\alpha {)}^3$ $=-3\alpha \beta \gamma$ [ Since, $\alpha +\beta +\gamma =0$ ]

$\Rightarrow$ $(\alpha +\beta {)}^3+(\beta +\gamma {)}^3+(\gamma +\alpha {)}^3$ $=-3(-251)=753$