Answer :
We start with the equation
[tex]$$
x^6 - 7x^3 - 8 = 0.
$$[/tex]
A useful strategy is to make the substitution
[tex]$$
y = x^3.
$$[/tex]
Then the equation becomes
[tex]$$
y^2 - 7y - 8 = 0.
$$[/tex]
### Step 1. Solve the Quadratic Equation
The quadratic
[tex]$$
y^2 - 7y - 8 = 0
$$[/tex]
can be factored as
[tex]$$
(y - 8)(y + 1) = 0.
$$[/tex]
Thus, we have
[tex]$$
y = 8 \quad \text{or} \quad y = -1.
$$[/tex]
Recall that [tex]$y = x^3$[/tex]. This gives us two separate equations:
1. [tex]$$x^3 = 8,$$[/tex]
2. [tex]$$x^3 = -1.$$[/tex]
### Step 2. Solve [tex]$x^3 = 8$[/tex]
The cube roots of 8 are found by writing
[tex]$$
8 = 2^3,
$$[/tex]
so one root is
[tex]$$
x = 2.
$$[/tex]
The other cube roots are the complex numbers given by
[tex]$$
x = 2 e^{\frac{2\pi i}{3}} \quad \text{and} \quad x = 2 e^{\frac{4\pi i}{3}}.
$$[/tex]
### Step 3. Solve [tex]$x^3 = -1$[/tex]
Expressing [tex]$-1$[/tex] in polar form, we write
[tex]$$
-1 = e^{i\pi}
$$[/tex]
(with the understanding that [tex]$-1 = e^{i(\pi + 2\pi k)}$[/tex] for any integer [tex]$k$[/tex]). The cube roots are then given by
[tex]$$
x = e^{\frac{i(\pi + 2\pi k)}{3}},\quad k = 0, 1, 2.
$$[/tex]
Evaluating for each [tex]$k$[/tex]:
- For [tex]$k = 0$[/tex]:
[tex]$$
x = e^{\frac{i\pi}{3}} = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2},
$$[/tex]
- For [tex]$k = 1$[/tex]:
[tex]$$
x = e^{\frac{i(\pi+2\pi)}{3}} = e^{i\pi} = -1,
$$[/tex]
- For [tex]$k = 2$[/tex]:
[tex]$$
x = e^{\frac{i(\pi+4\pi)}{3}} = e^{\frac{i5\pi}{3}} = \cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}.
$$[/tex]
### Final Answer
Combining the results from both cases, the complete set of solutions for the equation
[tex]$$
x^6 - 7x^3 - 8 = 0
$$[/tex]
is:
- From [tex]$x^3 = 8$[/tex]:
[tex]$$
x = 2,\quad x = 2 e^{\frac{2\pi i}{3}},\quad x = 2 e^{\frac{4\pi i}{3}};
$$[/tex]
- From [tex]$x^3 = -1$[/tex]:
[tex]$$
x = -1,\quad x = e^{\frac{i\pi}{3}} = \frac{1}{2} + i\frac{\sqrt{3}}{2},\quad x = e^{\frac{i5\pi}{3}} = \frac{1}{2} - i\frac{\sqrt{3}}{2}.
$$[/tex]
Thus, the six solutions to the equation are
[tex]$$
\boxed{
\begin{array}{l}
x = 2,\quad x = 2 e^{\frac{2\pi i}{3}},\quad x = 2 e^{\frac{4\pi i}{3}},\\[1mm]
x = -1,\quad x = \frac{1}{2} + i\frac{\sqrt{3}}{2},\quad x = \frac{1}{2} - i\frac{\sqrt{3}}{2}.
\end{array}
}
$$[/tex]
[tex]$$
x^6 - 7x^3 - 8 = 0.
$$[/tex]
A useful strategy is to make the substitution
[tex]$$
y = x^3.
$$[/tex]
Then the equation becomes
[tex]$$
y^2 - 7y - 8 = 0.
$$[/tex]
### Step 1. Solve the Quadratic Equation
The quadratic
[tex]$$
y^2 - 7y - 8 = 0
$$[/tex]
can be factored as
[tex]$$
(y - 8)(y + 1) = 0.
$$[/tex]
Thus, we have
[tex]$$
y = 8 \quad \text{or} \quad y = -1.
$$[/tex]
Recall that [tex]$y = x^3$[/tex]. This gives us two separate equations:
1. [tex]$$x^3 = 8,$$[/tex]
2. [tex]$$x^3 = -1.$$[/tex]
### Step 2. Solve [tex]$x^3 = 8$[/tex]
The cube roots of 8 are found by writing
[tex]$$
8 = 2^3,
$$[/tex]
so one root is
[tex]$$
x = 2.
$$[/tex]
The other cube roots are the complex numbers given by
[tex]$$
x = 2 e^{\frac{2\pi i}{3}} \quad \text{and} \quad x = 2 e^{\frac{4\pi i}{3}}.
$$[/tex]
### Step 3. Solve [tex]$x^3 = -1$[/tex]
Expressing [tex]$-1$[/tex] in polar form, we write
[tex]$$
-1 = e^{i\pi}
$$[/tex]
(with the understanding that [tex]$-1 = e^{i(\pi + 2\pi k)}$[/tex] for any integer [tex]$k$[/tex]). The cube roots are then given by
[tex]$$
x = e^{\frac{i(\pi + 2\pi k)}{3}},\quad k = 0, 1, 2.
$$[/tex]
Evaluating for each [tex]$k$[/tex]:
- For [tex]$k = 0$[/tex]:
[tex]$$
x = e^{\frac{i\pi}{3}} = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2},
$$[/tex]
- For [tex]$k = 1$[/tex]:
[tex]$$
x = e^{\frac{i(\pi+2\pi)}{3}} = e^{i\pi} = -1,
$$[/tex]
- For [tex]$k = 2$[/tex]:
[tex]$$
x = e^{\frac{i(\pi+4\pi)}{3}} = e^{\frac{i5\pi}{3}} = \cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}.
$$[/tex]
### Final Answer
Combining the results from both cases, the complete set of solutions for the equation
[tex]$$
x^6 - 7x^3 - 8 = 0
$$[/tex]
is:
- From [tex]$x^3 = 8$[/tex]:
[tex]$$
x = 2,\quad x = 2 e^{\frac{2\pi i}{3}},\quad x = 2 e^{\frac{4\pi i}{3}};
$$[/tex]
- From [tex]$x^3 = -1$[/tex]:
[tex]$$
x = -1,\quad x = e^{\frac{i\pi}{3}} = \frac{1}{2} + i\frac{\sqrt{3}}{2},\quad x = e^{\frac{i5\pi}{3}} = \frac{1}{2} - i\frac{\sqrt{3}}{2}.
$$[/tex]
Thus, the six solutions to the equation are
[tex]$$
\boxed{
\begin{array}{l}
x = 2,\quad x = 2 e^{\frac{2\pi i}{3}},\quad x = 2 e^{\frac{4\pi i}{3}},\\[1mm]
x = -1,\quad x = \frac{1}{2} + i\frac{\sqrt{3}}{2},\quad x = \frac{1}{2} - i\frac{\sqrt{3}}{2}.
\end{array}
}
$$[/tex]
