Answer :
To solve the equation [tex]\(x^6 - 7x^3 - 8 = 0\)[/tex], let's go through a step-by-step process:
1. Substitution: Begin by noticing that the equation is in terms of [tex]\(x^6\)[/tex] and [tex]\(x^3\)[/tex]. To simplify, let's make a substitution: set [tex]\(y = x^3\)[/tex]. This means [tex]\(x^6 = (x^3)^2 = y^2\)[/tex].
2. Rewriting the equation: Substitute [tex]\(y\)[/tex] into the equation:
[tex]\[
y^2 - 7y - 8 = 0
\][/tex]
3. Solving the quadratic equation: Now, solve the quadratic equation [tex]\(y^2 - 7y - 8 = 0\)[/tex].
Use 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]:
[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]
Thus, [tex]\(y = 8\)[/tex] or [tex]\(y = -1\)[/tex].
4. Recovering [tex]\(x\)[/tex]:
- For [tex]\(y = 8\)[/tex], equating back to [tex]\(x^3 = 8\)[/tex], we solve for [tex]\(x\)[/tex]:
[tex]\[
x^3 = 8 \implies x = 2
\][/tex]
- For [tex]\(y = -1\)[/tex], equating back to [tex]\(x^3 = -1\)[/tex], we solve for [tex]\(x\)[/tex]:
[tex]\[
x^3 = -1 \implies x = -1
\][/tex]
5. Complex Solutions: Since [tex]\(x\)[/tex] is a cubic root, there are more possible complex roots.
- The cube roots of [tex]\(8\)[/tex] are: [tex]\(2\)[/tex], and the complex numbers [tex]\(\frac{1}{2} - \frac{\sqrt{3}}{2}i\)[/tex] and [tex]\(\frac{1}{2} + \frac{\sqrt{3}}{2}i\)[/tex].
- The cube roots of [tex]\(-1\)[/tex] are: [tex]\(-1\)[/tex], and the complex numbers: [tex]\(-1 - \sqrt{3}i\)[/tex] and [tex]\(-1 + \sqrt{3}i\)[/tex].
So, the solutions for the equation [tex]\(x^6 - 7x^3 - 8 = 0\)[/tex] are:
[tex]\[
x = -1, \quad x = 2, \quad x = -1 - \sqrt{3}i, \quad x = -1 + \sqrt{3}i, \quad x = \frac{1}{2} - \frac{\sqrt{3}}{2}i, \quad x = \frac{1}{2} + \frac{\sqrt{3}}{2}i
\][/tex]
1. Substitution: Begin by noticing that the equation is in terms of [tex]\(x^6\)[/tex] and [tex]\(x^3\)[/tex]. To simplify, let's make a substitution: set [tex]\(y = x^3\)[/tex]. This means [tex]\(x^6 = (x^3)^2 = y^2\)[/tex].
2. Rewriting the equation: Substitute [tex]\(y\)[/tex] into the equation:
[tex]\[
y^2 - 7y - 8 = 0
\][/tex]
3. Solving the quadratic equation: Now, solve the quadratic equation [tex]\(y^2 - 7y - 8 = 0\)[/tex].
Use 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]:
[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]
Thus, [tex]\(y = 8\)[/tex] or [tex]\(y = -1\)[/tex].
4. Recovering [tex]\(x\)[/tex]:
- For [tex]\(y = 8\)[/tex], equating back to [tex]\(x^3 = 8\)[/tex], we solve for [tex]\(x\)[/tex]:
[tex]\[
x^3 = 8 \implies x = 2
\][/tex]
- For [tex]\(y = -1\)[/tex], equating back to [tex]\(x^3 = -1\)[/tex], we solve for [tex]\(x\)[/tex]:
[tex]\[
x^3 = -1 \implies x = -1
\][/tex]
5. Complex Solutions: Since [tex]\(x\)[/tex] is a cubic root, there are more possible complex roots.
- The cube roots of [tex]\(8\)[/tex] are: [tex]\(2\)[/tex], and the complex numbers [tex]\(\frac{1}{2} - \frac{\sqrt{3}}{2}i\)[/tex] and [tex]\(\frac{1}{2} + \frac{\sqrt{3}}{2}i\)[/tex].
- The cube roots of [tex]\(-1\)[/tex] are: [tex]\(-1\)[/tex], and the complex numbers: [tex]\(-1 - \sqrt{3}i\)[/tex] and [tex]\(-1 + \sqrt{3}i\)[/tex].
So, the solutions for the equation [tex]\(x^6 - 7x^3 - 8 = 0\)[/tex] are:
[tex]\[
x = -1, \quad x = 2, \quad x = -1 - \sqrt{3}i, \quad x = -1 + \sqrt{3}i, \quad x = \frac{1}{2} - \frac{\sqrt{3}}{2}i, \quad x = \frac{1}{2} + \frac{\sqrt{3}}{2}i
\][/tex]
