The number of distinct real roots of the equation x ^ 5 ( x ^ 3 - x...

Given,

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

Now on rearranging we get

$⇒ {x}^{8} - {x}^{7} - {x}^{6} + {x}^{5} + 3{x}^{4} - 4{x}^{3} - 2{x}^{2} + 4x - 1 = 0$

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

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

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

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

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

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

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

Hence $3$ real distinct roots.