The number of distinct real roots of the equation x 5 x 3 - x 2 - x + 1 + x 3 x 3 - 4 x 2 - 2 x + 4 - 1 = 0 is

Given,

$x^5\left(x^3-x^2-x+1\right)+x\left(3x^3-4x^2-2x+4\right)-1=0$

Now on rearranging we get

$\Rightarrow x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1=0$

$\Rightarrow x^7\left(x-1\right)-x^5\left(x-1\right)+3x^3\left(x-1\right)-x\left(x^2-1\right)-2x\left(x-1\right)+x-1=0$

$\Rightarrow \left(x-1\right)\left(x^7-x^5+3x^3-x\left(x+1\right)-2x+1\right)=0$

$\Rightarrow \left(x-1\right)\left(x^5\left(x^2-1\right)+3x\left(x^2-1\right)-1\left(x^2-1\right)\right)=0$

$\Rightarrow \left(x^2-1\right)\left(x-1\right)\left(x^5+3x-1\right)=0$

From above equation it is clearly visible that the root are $x=\pm 1$ and for $x^5+3x-1$

Let $f\left(x\right)=x^5+3x-1$, $f^{\text{'}}\left(x\right)=5x^4+3$

$f^{\text{'}}\left(x\right)>0\forall x\in R$, hence it is monotonic in nature so it will have only one root other than $1 \& -1$.

Hence $3$ real distinct roots.