The number of distinct real roots of the equation $$x^5(x^3 - x^2 - x + 1) + x(3x^3 - 4x^2 - 2x + 4) - 1 = 0$$ is ______.

Given,

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

First factor each bracket.

For

$$x^3-x^2-x+1,$$

group terms:

$$x^2(x-1)-1(x-1)$$

$$=(x-1)(x^2-1)$$

$$=(x-1)^2(x+1)$$

Now factor

$$3x^3-4x^2-2x+4$$

Grouping,

$$x^2(3x-4)-2(3x-4)$$

$$=(3x-4)(x^2-2)$$

Hence equation becomes

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

Now expand completely:

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

Now check rational roots.

Substituting

$$x=1,$$

$$1-1-1+1+3-4-2+4-1=0$$

Hence,

$$(x-1)$$

is a factor.

Using synthetic division:

$$x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1$$

$$=(x-1)(x^7-x^5+3x^3-x^2-3x+1)$$

Now check

$$x=-1$$

in the quotient:

$$-1+1+3-1+3+1=0$$

Hence,

$$(x+1)$$

is also a factor.

Dividing again,

$$=(x-1)(x+1)(x^6-x^5+x^4-x^3+2x^2-3x+1)$$

Now test

$$x=1$$

again:

$$1-1+1-1+2-3+1=0$$

Hence another factor

$$(x-1)$$

appears.

Therefore,

$$=(x-1)^2(x+1)(x^5+x^3+2x-1)$$

Now consider

$$x^5+x^3+2x-1$$

Its derivative is

$$5x^4+3x^2+2>0$$

Hence this function is strictly increasing and has exactly one real root.

Thus real roots are:

- $$x=1$$

- $$x=-1$$

- one real root from $$x^5+x^3+2x-1=0$$

Therefore, number of distinct real roots is

$$\boxed{3}$$.