Forth root of the equation

Taking the 4th root from R.H.S. ( Right hand side) of equation we get

$(x + 1)^{4}$ = $x^{4}$ + 4 $x^{3}$ - x

or, $(x + 1)^{2}$ $(x + 1)^{2}$ = $x^{4}$ + 4 $x^{3}$ - x

By using the formula of $(a + b)^{2}$

( $x^{2}$ + 2x + 1)( $x^{2}$ + 2x + 1) = $x^{4}$ + 4 $x^{3}$ - x

or, $x^{4}$ + 2 $x^{3}$ + $x^{2}$ + 2 $x^{3}$ + 4 $x^{2}$ + 2x + $x^{2}$ + 2x + 1 = $x^{4}$ + 4 $x^{3}$ - x

or, $x^{4}$ + 4 $x^{3}$ + $x^{2}$ + 5 $x^{2}$ + 4x + 1 = $x^{4}$ + 4 $x^{3}$ - x

or, $x^{2}$ + 5 $x^{2}$ + 4x + 1 +x = 0

or, 6 $x^{2}$ + 5x + 1 = 0

or, 6 $x^{2}$ + 3x + 2x + 1 = 0

or, (3x + 1)(2x + 1) = 0

Either,(3x + 1) = 0-------------------------------------------------------( i )

or, (2x + 1) = 0-------------------------------------------------------( ii )

From equation ( i )

(3x + 1) = 0

or, 3x = -1

or, x = - 1/3

From equation ( ii )

(2x + 1) = 0

or, 2x = -1

or, x = - 1/2

Hence, x = $\boxed{ - 1/2 and - 1/3}$