Answer :
To solve the equation [tex]\( x^6 - 7x^3 - 8 = 0 \)[/tex], we can start by recognizing that it's a polynomial equation with [tex]\( x^6 \)[/tex] as the highest degree. Given the hint that there are 6 solutions, we'll look to find these solutions which may include real and complex numbers.
### Step 1: Substitution
Since the highest power is [tex]\( x^6 \)[/tex], let's make a substitution to simplify our work. Let [tex]\( y = x^3 \)[/tex]. This means that [tex]\( x^6 = (x^3)^2 = y^2 \)[/tex].
So, the equation [tex]\( x^6 - 7x^3 - 8 = 0 \)[/tex] becomes:
[tex]\[ y^2 - 7y - 8 = 0 \][/tex]
### Step 2: Solve the Quadratic Equation
Now, we need to solve the quadratic equation in terms of [tex]\( y \)[/tex]:
[tex]\[ y^2 - 7y - 8 = 0 \][/tex]
We can use the quadratic formula where [tex]\( a = 1 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = -8 \)[/tex]:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute the values:
[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 two solutions for [tex]\( y \)[/tex]:
1. [tex]\( y = \frac{7 + 9}{2} = 8 \)[/tex]
2. [tex]\( y = \frac{7 - 9}{2} = -1 \)[/tex]
### Step 3: Back-substitution to Find [tex]\( x \)[/tex]
Remember, [tex]\( y = x^3 \)[/tex], so we substitute back to find the values of [tex]\( x \)[/tex]:
1. For [tex]\( y = 8 \)[/tex]:
[tex]\[ x^3 = 8 \][/tex]
[tex]\[ x = \sqrt[3]{8} = 2 \][/tex]
2. For [tex]\( y = -1 \)[/tex]:
[tex]\[ x^3 = -1 \][/tex]
[tex]\[ x = \sqrt[3]{-1} = -1 \][/tex]
### Step 4: Complex Solutions
The solutions above are real and straightforward. Since we expect 6 solutions total (including complex numbers), we examine the complex roots of each [tex]\( y \)[/tex]-value:
For [tex]\( y = 8 \)[/tex], the complex cube roots also come into play. For [tex]\( x^3 = 8 \)[/tex], aside from the real solution [tex]\( x = 2 \)[/tex], the complex roots are calculated using the formula for complex cube roots, resulting in solutions:
[tex]\[ x = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i \][/tex]
For [tex]\( y = -1 \)[/tex], aside from the real solution [tex]\( x = -1 \)[/tex], similar complex calculations will give:
[tex]\[ x = -1 \pm \sqrt{3}i \][/tex]
### Conclusion:
Altogether, the complete set of solutions where the polynomial equation holds true includes both the real and complex numbers:
[tex]\[ x = -1, 2, -1 - \sqrt{3}i, -1 + \sqrt{3}i, \frac{1}{2} - \frac{\sqrt{3}}{2}i, \frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
These are the six solutions to the given equation [tex]\( x^6 - 7x^3 - 8 = 0 \)[/tex].
### Step 1: Substitution
Since the highest power is [tex]\( x^6 \)[/tex], let's make a substitution to simplify our work. Let [tex]\( y = x^3 \)[/tex]. This means that [tex]\( x^6 = (x^3)^2 = y^2 \)[/tex].
So, the equation [tex]\( x^6 - 7x^3 - 8 = 0 \)[/tex] becomes:
[tex]\[ y^2 - 7y - 8 = 0 \][/tex]
### Step 2: Solve the Quadratic Equation
Now, we need to solve the quadratic equation in terms of [tex]\( y \)[/tex]:
[tex]\[ y^2 - 7y - 8 = 0 \][/tex]
We can use the quadratic formula where [tex]\( a = 1 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = -8 \)[/tex]:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute the values:
[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 two solutions for [tex]\( y \)[/tex]:
1. [tex]\( y = \frac{7 + 9}{2} = 8 \)[/tex]
2. [tex]\( y = \frac{7 - 9}{2} = -1 \)[/tex]
### Step 3: Back-substitution to Find [tex]\( x \)[/tex]
Remember, [tex]\( y = x^3 \)[/tex], so we substitute back to find the values of [tex]\( x \)[/tex]:
1. For [tex]\( y = 8 \)[/tex]:
[tex]\[ x^3 = 8 \][/tex]
[tex]\[ x = \sqrt[3]{8} = 2 \][/tex]
2. For [tex]\( y = -1 \)[/tex]:
[tex]\[ x^3 = -1 \][/tex]
[tex]\[ x = \sqrt[3]{-1} = -1 \][/tex]
### Step 4: Complex Solutions
The solutions above are real and straightforward. Since we expect 6 solutions total (including complex numbers), we examine the complex roots of each [tex]\( y \)[/tex]-value:
For [tex]\( y = 8 \)[/tex], the complex cube roots also come into play. For [tex]\( x^3 = 8 \)[/tex], aside from the real solution [tex]\( x = 2 \)[/tex], the complex roots are calculated using the formula for complex cube roots, resulting in solutions:
[tex]\[ x = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i \][/tex]
For [tex]\( y = -1 \)[/tex], aside from the real solution [tex]\( x = -1 \)[/tex], similar complex calculations will give:
[tex]\[ x = -1 \pm \sqrt{3}i \][/tex]
### Conclusion:
Altogether, the complete set of solutions where the polynomial equation holds true includes both the real and complex numbers:
[tex]\[ x = -1, 2, -1 - \sqrt{3}i, -1 + \sqrt{3}i, \frac{1}{2} - \frac{\sqrt{3}}{2}i, \frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
These are the six solutions to the given equation [tex]\( x^6 - 7x^3 - 8 = 0 \)[/tex].
