The number of distinct real roots of the equation $x^{5}\left(x^{3}-x^{2}-x+1\right)+x\left(3 x^{3}-4 x^{2}-2 x+4\right)-1=0$ is ______________.

<p>${x^8} - {x^7} - {x^6} + {x^5} + 3{x^4} - 4{x^3} - 2{x^2} + 4x - 1 = 0$</p> <p>$$ \Rightarrow {x^7}(x - 1) - {x^5}(x - 1) + 3{x^3}(x - 1) - x({x^2} - 1) + 2x(1 - x) + (x - 1) = 0$$</p> <p>$\Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - x(x + 1) - 2x + 1) = 0$</p> <p>$\Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - {x^2} - 3x + 1) = 0$</p> <p>$\Rightarrow (x - 1)({x^5}({x^2} - 1) + 3x({x^2} - 1) - 1({x^2} - 1)) = 0$</p> <p>$\Rightarrow (x - 1)({x^2} - 1)({x^5} + 3x - 1) = 0$</p> <p>$\therefore$ $x = \, \pm \,1$ are roots of above equation and ${x^5} + 3x - 1$ is a monotonic term hence vanishs at exactly one value of x other than 1 or $-$1.</p> <p>$\therefore$ 3 real roots.</p>