-1 satisfies the given equation. Thus,
$7x^4 (x+1) - 7x^3 (x+1) + 7x^2 (x+1) -7x(x+1)+8(x+1)=0$
$\Rightarrow (x+1) (7x^4 - 7x^3 +7x^2 -7x +8) = 0$
Consider, $7x^4 - 7x^3 +7x^2 -7x +8 = 0$
Or, $7(x^4 - x^3 +x^2 -x) +8 = 0$
Let us consider III cases,
I. x < 0
Since $x^3$ and x both terms are negative, the left side is always +ve and cannot be 0. So, no solution is possible in this case.
II. x > 1
$x^4 > x^3$ and $x^2 > x$. So, left side is always +ve and cannot be 0. So, no solution is possible in this case.
III 0 < x < 1
$x^4 - x^3$ is a -ve decimal and so is $x^2 - x$. So, left side will be slightly smaller than 8 but cannot be 0. No solution in this case also.
As can be seen x cannot be 0. x cannot be 1 either.
The only solution possible is -1 for the original equation.
