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

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

Let $\alpha=a+b,\beta=ab$
$a^3(a+1)=1$ and $b^3(b+1)=1$ which gives
$1=(ab)^3(a+1)(b+1)$
$1=(ab)^3(ab+a+b+1)=\beta^3(\beta+\alpha+1)$
Also,$a^4+a^3=b^4+b^3$
$(a^3+a^2)+(b^3+b^2)+ab(a+b)+ab=0$
$\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.