behöver hjälp med ett tal: 3=-x+x^-2

Solve for x over the real numbers:
3 = 1/x^2-x
Reverse the equality in 3 = 1/x^2-x in order to isolate x to the left hand side.
3 = 1/x^2-x is equivalent to 1/x^2-x = 3:
1/x^2-x = 3
Write the left hand side as a single fraction.
Bring 1/x^2-x together using the common denominator x^2:
(1-x^3)/x^2 = 3
Multiply both sides by a polynomial to clear fractions.
Multiply both sides by x^2:
1-x^3 = 3 x^2
Move everything to the left hand side.
Subtract 3 x^2 from both sides:
-x^3-3 x^2+1 = 0
Divide both sides by the sign of the leading coefficient of -x^3-3 x^2+1.
Multiply both sides by -1:
x^3+3 x^2-1 = 0
Look for a simple substitution that eliminates the quadratic term of x^3+3 x^2-1.
Eliminate the quadratic term by substituting y = x+1:
-1+3 (y-1)^2+(y-1)^3 = 0
Write the cubic polynomial on the left hand side in standard form.
Expand out terms of the left hand side:
y^3-3 y+1 = 0
Change coordinates by substituting y = lambda/z+z, where lambda is a constant value that will be determined later.
If y = lambda/z+z then z = 1/2 (y+sqrt(y^2-4 lambda)) which will be used during back substitution:
1-3 (z+lambda/z)+(z+lambda/z)^3 = 0
Transform the rational equation into a polynomial equation.
Multiply both sides by z^3 and collect in terms of z:
z^6+z^4 (3 lambda-3)+z^3+z^2 (3 lambda^2-3 lambda)+lambda^3 = 0
Find an appropriate value for lambda in order to make the coefficients of z^2 and z^4 both zero.
Substitute lambda = 1 and then u = z^3, yielding a quadratic equation in the variable u:
u^2+u+1 = 0
Solve for u.
Find the positive solution to the quadratic equation:
u = 1/2 i (i+sqrt(3))
Perform back substitution on u = 1/2 i (i+sqrt(3)).
Substitute back for u = z^3:
z^3 = 1/2 i (i+sqrt(3))
Move everything to the left hand side.
Subtract 1/2 i (i+sqrt(3)) from both sides:
1/2 (-i) (i+sqrt(3))+z^3 = 0
Divide by an appropriate factor to make the constant term 1.
Divide both sides by 1/2 i (i+sqrt(3)):
-((2 i) z^3)/(i+sqrt(3))-1 = 0
Write -(2 i z^3)/(i+sqrt(3)) with a common power.
-((2 i) z^3)/(i+sqrt(3))-1 = z^3 (-((-1)^(5/6))/((1/2 (i+sqrt(3)))^(1/3)))^3-1 = (-((-1)^(5/6) z)/((1/2 (i+sqrt(3)))^(1/3)))^3-1:
(-((-1)^(5/6) z)/((1/2 (i+sqrt(3)))^(1/3)))^3-1 = 0
Simplify (-((-1)^(5/6) z)/(1/2 (i+sqrt(3)))^(1/3))^3-1 = 0 by making a substitution.
Substitute v = -((-1)^(5/6) 2^(1/3) z)/(i+sqrt(3))^(1/3):
v^3-1 = 0
Solve for v.
By definition, the 3^rd roots of unity are the roots of v^3-1 = 0:
v = 1 or v = -(-1)^(1/3) or v = (-1)^(2/3)
Look at the first equation: Perform back substitution on v = 1.
Substitute back for v = -((-1)^(5/6) 2^(1/3) z)/(i+sqrt(3))^(1/3):
-((-1)^(5/6) 2^(1/3) z)/(i+sqrt(3))^(1/3) = 1 or v = -(-1)^(1/3) or v = (-1)^(2/3)
Solve for z.
Divide both sides by -((-1)^(5/6) 2^(1/3))/(i+sqrt(3))^(1/3):
z = ((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or v = -(-1)^(1/3) or v = (-1)^(2/3)
Perform back substitution on z = ((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3).
Substitute back for z = y/2+sqrt(y^2-4)/2:
y/2+sqrt(y^2-4)/2 = ((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or v = -(-1)^(1/3) or v = (-1)^(2/3)
Solve for y.
Solve the radical equation in y:
y = ((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3)-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3) or v = -(-1)^(1/3) or v = (-1)^(2/3)
Perform back substitution on y = ((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3)-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3).
Substitute back for y = x+1:
x+1 = ((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3)-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3) or v = -(-1)^(1/3) or v = (-1)^(2/3)
Solve for x.
Subtract 1 from both sides:
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or v = -(-1)^(1/3) or v = (-1)^(2/3)
Look at the second equation: Perform back substitution on v = -(-1)^(1/3).
Substitute back for v = -((-1)^(5/6) 2^(1/3) z)/(i+sqrt(3))^(1/3):
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or -((-1)^(5/6) 2^(1/3) z)/(i+sqrt(3))^(1/3) = -(-1)^(1/3) or v = (-1)^(2/3)
Solve for z.
Divide both sides by -((-1)^(5/6) 2^(1/3))/(i+sqrt(3))^(1/3):
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or z = -(i (i+sqrt(3))^(1/3))/2^(1/3) or v = (-1)^(2/3)
Perform back substitution on z = -(i (i+sqrt(3))^(1/3))/2^(1/3).
Substitute back for z = y/2+sqrt(y^2-4)/2:
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or y/2+sqrt(y^2-4)/2 = -(i (i+sqrt(3))^(1/3))/2^(1/3) or v = (-1)^(2/3)
Solve for y.
Solve the radical equation in y:
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or y = i/(1/2 (i+sqrt(3)))^(1/3)-i (1/2 (i+sqrt(3)))^(1/3) or v = (-1)^(2/3)
Perform back substitution on y = i/(1/2 (i+sqrt(3)))^(1/3)-i (1/2 (i+sqrt(3)))^(1/3).
Substitute back for y = x+1:
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or x+1 = i/(1/2 (i+sqrt(3)))^(1/3)-i (1/2 (i+sqrt(3)))^(1/3) or v = (-1)^(2/3)
Solve for x.
Subtract 1 from both sides:
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or x = -1+i/(1/2 (i+sqrt(3)))^(1/3)-i (1/2 (i+sqrt(3)))^(1/3) or v = (-1)^(2/3)
Look at the third equation: Perform back substitution on v = (-1)^(2/3).
Substitute back for v = -((-1)^(5/6) 2^(1/3) z)/(i+sqrt(3))^(1/3):
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or x = -1+i/(1/2 (i+sqrt(3)))^(1/3)-i (1/2 (i+sqrt(3)))^(1/3) or -((-1)^(5/6) 2^(1/3) z)/(i+sqrt(3))^(1/3) = (-1)^(2/3)
Solve for z.
Divide both sides by -((-1)^(5/6) 2^(1/3))/(i+sqrt(3))^(1/3):
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or x = -1+i/(1/2 (i+sqrt(3)))^(1/3)-i (1/2 (i+sqrt(3)))^(1/3) or z = ((-1)^(5/6) (i+sqrt(3))^(1/3))/2^(1/3)
Perform back substitution on z = ((-1)^(5/6) (i+sqrt(3))^(1/3))/2^(1/3).
Substitute back for z = y/2+sqrt(y^2-4)/2:
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or x = -1+i/(1/2 (i+sqrt(3)))^(1/3)-i (1/2 (i+sqrt(3)))^(1/3) or y/2+sqrt(y^2-4)/2 = ((-1)^(5/6) (i+sqrt(3))^(1/3))/2^(1/3)
Solve for y.
Solve the radical equation in y:
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or x = -1+i/(1/2 (i+sqrt(3)))^(1/3)-i (1/2 (i+sqrt(3)))^(1/3) or y = (-1)^(5/6) (1/2 (i+sqrt(3)))^(1/3)-(-1)^(1/6)/(1/2 (i+sqrt(3)))^(1/3)
Perform back substitution on y = (-1)^(5/6) (1/2 (i+sqrt(3)))^(1/3)-(-1)^(1/6)/(1/2 (i+sqrt(3)))^(1/3).
Substitute back for y = x+1:
x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or x = -1+i/(1/2 (i+sqrt(3)))^(1/3)-i (1/2 (i+sqrt(3)))^(1/3) or x+1 = (-1)^(5/6) (1/2 (i+sqrt(3)))^(1/3)-(-1)^(1/6)/(1/2 (i+sqrt(3)))^(1/3)
Solve for x.
Subtract 1 from both sides:
Answer: |
| x = -1-(-1)^(5/6)/(1/2 (i+sqrt(3)))^(1/3)+((-1)^(1/6) (i+sqrt(3))^(1/3))/2^(1/3) or x = -1+i/(1/2 (i+sqrt(3)))^(1/3)-i (1/2 (i+sqrt(3)))^(1/3) or x = -1-(-1)^(1/6)/(1/2 (i+sqrt(3)))^(1/3)+(-1)^(5/6) (1/2 (i+sqrt(3)))^(1/3)