Answer :
To solve the equation [tex]\(x^6 - 7x^3 - 8 = 0\)[/tex], we can start by substituting a new variable to simplify the equation. Let's set [tex]\(y = x^3\)[/tex]. This means [tex]\(y^2 = x^6\)[/tex]. Substituting these into the equation, we get:
[tex]\[ y^2 - 7y - 8 = 0 \][/tex]
This is a quadratic equation in terms of [tex]\(y\)[/tex]. We can solve it using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = -7\)[/tex], and [tex]\(c = -8\)[/tex]. Plugging these values into the formula gives:
[tex]\[ y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{7 \pm \sqrt{49 + 32}}{2} \][/tex]
[tex]\[ y = \frac{7 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ y = \frac{7 \pm 9}{2} \][/tex]
This gives us the solutions:
1. [tex]\( y = \frac{7 + 9}{2} = 8 \)[/tex]
2. [tex]\( y = \frac{7 - 9}{2} = -1 \)[/tex]
Since [tex]\(y = x^3\)[/tex], we now have two separate equations to solve for [tex]\(x\)[/tex]:
1. [tex]\( x^3 = 8 \)[/tex]
2. [tex]\( x^3 = -1 \)[/tex]
Solving [tex]\(x^3 = 8\)[/tex]:
The solutions for this are the cube roots of 8. The real cube root of 8 is 2, but complex numbers might also serve as solutions using the properties of complex numbers and roots of unity. However, here we focus on the real numbers:
- [tex]\(x = 2\)[/tex]
Solving [tex]\(x^3 = -1\)[/tex]:
The solutions for this are the cube roots of -1:
- The principal real cube root of -1 is -1.
However, there are other complex roots for this equation, which come from the property of complex numbers:
- [tex]\(x = -1\)[/tex]
- The complex roots, given in the result as [tex]\(-1 - \sqrt{3}i\)[/tex] and [tex]\(-1 + \sqrt{3}i\)[/tex], are also considered.
Final Roots:
Combining real and complex solutions for the initial polynomial equation [tex]\(x^6 - 7x^3 - 8 = 0\)[/tex], we have:
- [tex]\(x = -1\)[/tex]
- [tex]\(x = 2\)[/tex]
- [tex]\(x = -1 - \sqrt{3}i\)[/tex]
- [tex]\(x = -1 + \sqrt{3}i\)[/tex]
- [tex]\(x = \frac{1}{2} - \frac{\sqrt{3}}{2}i\)[/tex]
- [tex]\(x = \frac{1}{2} + \frac{\sqrt{3}}{2}i\)[/tex]
These are the complete set of solutions for the given equation.
[tex]\[ y^2 - 7y - 8 = 0 \][/tex]
This is a quadratic equation in terms of [tex]\(y\)[/tex]. We can solve it using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = -7\)[/tex], and [tex]\(c = -8\)[/tex]. Plugging these values into the formula gives:
[tex]\[ y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{7 \pm \sqrt{49 + 32}}{2} \][/tex]
[tex]\[ y = \frac{7 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ y = \frac{7 \pm 9}{2} \][/tex]
This gives us the solutions:
1. [tex]\( y = \frac{7 + 9}{2} = 8 \)[/tex]
2. [tex]\( y = \frac{7 - 9}{2} = -1 \)[/tex]
Since [tex]\(y = x^3\)[/tex], we now have two separate equations to solve for [tex]\(x\)[/tex]:
1. [tex]\( x^3 = 8 \)[/tex]
2. [tex]\( x^3 = -1 \)[/tex]
Solving [tex]\(x^3 = 8\)[/tex]:
The solutions for this are the cube roots of 8. The real cube root of 8 is 2, but complex numbers might also serve as solutions using the properties of complex numbers and roots of unity. However, here we focus on the real numbers:
- [tex]\(x = 2\)[/tex]
Solving [tex]\(x^3 = -1\)[/tex]:
The solutions for this are the cube roots of -1:
- The principal real cube root of -1 is -1.
However, there are other complex roots for this equation, which come from the property of complex numbers:
- [tex]\(x = -1\)[/tex]
- The complex roots, given in the result as [tex]\(-1 - \sqrt{3}i\)[/tex] and [tex]\(-1 + \sqrt{3}i\)[/tex], are also considered.
Final Roots:
Combining real and complex solutions for the initial polynomial equation [tex]\(x^6 - 7x^3 - 8 = 0\)[/tex], we have:
- [tex]\(x = -1\)[/tex]
- [tex]\(x = 2\)[/tex]
- [tex]\(x = -1 - \sqrt{3}i\)[/tex]
- [tex]\(x = -1 + \sqrt{3}i\)[/tex]
- [tex]\(x = \frac{1}{2} - \frac{\sqrt{3}}{2}i\)[/tex]
- [tex]\(x = \frac{1}{2} + \frac{\sqrt{3}}{2}i\)[/tex]
These are the complete set of solutions for the given equation.
