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Let $a,b$ be two distinct roots of $x^4+x^3-1=0$ and $p(x)=x^6+x^4+x^3-x^2-1$

$\begin{array}{1 1}(A)\;\text{a+b is a root of p(x)=0}\\(B)\;\text{ab is a root of p(x)=0}\\(C)\;\text{Both a+b and ab are root of p(x)=0}\\(D)\;\text{Neither a+b nor ab is a root of p(x)=0}\end{array} $

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Let $\alpha=a+b,\beta=ab$
$a^3(a+1)=1$ and $b^3(b+1)=1$ which gives
$\Rightarrow \large\frac{1}{a}+\frac{1}{b}$$+\alpha\beta+\beta=0$
$\Rightarrow \alpha(1+\beta^2)+\beta^2=0$
$\Rightarrow (\large\frac{1}{\beta^3}$$-\beta-1)(1+\beta^2)+\beta^2=0$
$\Rightarrow \beta^6+\beta^4+\beta^3-\beta^2-1=0$
Hence (B) is the correct answer.
answered Apr 16, 2014 by sreemathi.v

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