Answer :
To solve the equation [tex]\( x^6 - x^3 - 2 = 0 \)[/tex], follow these detailed steps:
1. Identify the type of equation:
This is a polynomial equation of degree 6.
2. Rewrite the equation:
Introduce a substitution to simplify the polynomial. Let [tex]\( y = x^3 \)[/tex]. Thus the equation becomes:
[tex]\[ y^2 - y - 2 = 0 \][/tex]
3. Solve the quadratic equation:
The equation [tex]\( y^2 - y - 2 = 0 \)[/tex] is a standard quadratic equation and can be solved using the quadratic formula. Recall that for a quadratic equation [tex]\( ay^2 + by + c = 0 \)[/tex], the solutions are given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -2 \)[/tex]. Plugging these values in:
[tex]\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \][/tex]
Simplify inside the square root:
[tex]\[ y = \frac{1 \pm \sqrt{1 + 8}}{2} \][/tex]
[tex]\[ y = \frac{1 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ y = \frac{1 \pm 3}{2} \][/tex]
Thus, the solutions for [tex]\( y \)[/tex] are:
[tex]\[ y = \frac{1 + 3}{2} = 2 \quad \text{and} \quad y = \frac{1 - 3}{2} = -1 \][/tex]
4. Back-substitute [tex]\( y = x^3 \)[/tex]:
We now have two equations to solve for [tex]\( x \)[/tex]:
[tex]\[ x^3 = 2 \quad \text{and} \quad x^3 = -1 \][/tex]
5. Solve [tex]\( x^3 = 2 \)[/tex]:
The solutions to [tex]\( x^3 = 2 \)[/tex] are the cube roots of 2:
[tex]\[ x = 2^{1/3}, \; \omega 2^{1/3}, \; \omega^2 2^{1/3} \][/tex]
where [tex]\( \omega \)[/tex] is a primitive cube root of unity satisfying [tex]\( \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \)[/tex] and [tex]\( \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \)[/tex].
Therefore:
[tex]\[ x = 2^{1/3}, \; -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
6. Solve [tex]\( x^3 = -1 \)[/tex]:
The solutions to [tex]\( x^3 = -1 \)[/tex] are the cube roots of -1:
[tex]\[ x = -1, \; \omega, \; \omega^2 \][/tex]
Note here [tex]\( \omega \)[/tex] and [tex]\( \omega^2 \)[/tex] remain the same as defined previously since they are roots of unity.
Therefore:
[tex]\[ x = -1, \; -\frac{1}{2} + \frac{\sqrt{3}}{2} i, \; -\frac{1}{2} - \frac{\sqrt{3}}{2} i \][/tex]
7. Compile all solutions:
Collecting all solutions from steps 5 and 6, we get:
[tex]\[ x = -1, 2^{1/3}, -\frac{1}{2} + \frac{\sqrt{3}}{2} i, -\frac{1}{2} - \frac{\sqrt{3}}{2} i, -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
Thus, the solutions to the equation [tex]\( x^6 - x^3 - 2 = 0 \)[/tex] are:
[tex]\[ -1, \; 2^{1/3}, \; -\frac{1}{2} + \frac{\sqrt{3}}{2} i, \; -\frac{1}{2} - \frac{\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
1. Identify the type of equation:
This is a polynomial equation of degree 6.
2. Rewrite the equation:
Introduce a substitution to simplify the polynomial. Let [tex]\( y = x^3 \)[/tex]. Thus the equation becomes:
[tex]\[ y^2 - y - 2 = 0 \][/tex]
3. Solve the quadratic equation:
The equation [tex]\( y^2 - y - 2 = 0 \)[/tex] is a standard quadratic equation and can be solved using the quadratic formula. Recall that for a quadratic equation [tex]\( ay^2 + by + c = 0 \)[/tex], the solutions are given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -2 \)[/tex]. Plugging these values in:
[tex]\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \][/tex]
Simplify inside the square root:
[tex]\[ y = \frac{1 \pm \sqrt{1 + 8}}{2} \][/tex]
[tex]\[ y = \frac{1 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ y = \frac{1 \pm 3}{2} \][/tex]
Thus, the solutions for [tex]\( y \)[/tex] are:
[tex]\[ y = \frac{1 + 3}{2} = 2 \quad \text{and} \quad y = \frac{1 - 3}{2} = -1 \][/tex]
4. Back-substitute [tex]\( y = x^3 \)[/tex]:
We now have two equations to solve for [tex]\( x \)[/tex]:
[tex]\[ x^3 = 2 \quad \text{and} \quad x^3 = -1 \][/tex]
5. Solve [tex]\( x^3 = 2 \)[/tex]:
The solutions to [tex]\( x^3 = 2 \)[/tex] are the cube roots of 2:
[tex]\[ x = 2^{1/3}, \; \omega 2^{1/3}, \; \omega^2 2^{1/3} \][/tex]
where [tex]\( \omega \)[/tex] is a primitive cube root of unity satisfying [tex]\( \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \)[/tex] and [tex]\( \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \)[/tex].
Therefore:
[tex]\[ x = 2^{1/3}, \; -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
6. Solve [tex]\( x^3 = -1 \)[/tex]:
The solutions to [tex]\( x^3 = -1 \)[/tex] are the cube roots of -1:
[tex]\[ x = -1, \; \omega, \; \omega^2 \][/tex]
Note here [tex]\( \omega \)[/tex] and [tex]\( \omega^2 \)[/tex] remain the same as defined previously since they are roots of unity.
Therefore:
[tex]\[ x = -1, \; -\frac{1}{2} + \frac{\sqrt{3}}{2} i, \; -\frac{1}{2} - \frac{\sqrt{3}}{2} i \][/tex]
7. Compile all solutions:
Collecting all solutions from steps 5 and 6, we get:
[tex]\[ x = -1, 2^{1/3}, -\frac{1}{2} + \frac{\sqrt{3}}{2} i, -\frac{1}{2} - \frac{\sqrt{3}}{2} i, -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
Thus, the solutions to the equation [tex]\( x^6 - x^3 - 2 = 0 \)[/tex] are:
[tex]\[ -1, \; 2^{1/3}, \; -\frac{1}{2} + \frac{\sqrt{3}}{2} i, \; -\frac{1}{2} - \frac{\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} + \frac{2^{1/3}\sqrt{3}}{2} i, \; -\frac{2^{1/3}}{2} - \frac{2^{1/3}\sqrt{3}}{2} i \][/tex]
