Sure, let's solve the equation [tex]\(2x^3 + 3x^3 = 5\)[/tex] step-by-step:
1. Combine Like Terms:
[tex]\[
2x^3 + 3x^3 = 5x^3
\][/tex]
So the given equation simplifies to:
[tex]\[
5x^3 = 5
\][/tex]
2. Isolate [tex]\(x^3\)[/tex]:
[tex]\[
x^3 = \frac{5}{5}
\][/tex]
Simplifying the fraction on the right side, we get:
[tex]\[
x^3 = 1
\][/tex]
3. Solve for [tex]\(x\)[/tex] (Find the cube roots of 1):
The equation [tex]\(x^3 = 1\)[/tex] has three solutions in the complex plane:
- Real Solution:
[tex]\[
x = 1
\][/tex]
- Complex Conjugate Pair Solutions:
These solutions arise from the roots of unity, specifically the cube roots of 1 (other than the real root):
- The roots of unity can be expressed in the form [tex]\(\text{cis} \theta\)[/tex] or [tex]\(e^{i\theta}\)[/tex], where [tex]\(\theta\)[/tex] is [tex]\(\frac{2\pi}{3}\)[/tex] radians for the non-real roots.
Therefore, the other two solutions are:
[tex]\[
x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \quad \text{and} \quad x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i
\][/tex]
Thus, the complete set of solutions to the equation [tex]\(2x^3 + 3x^3 = 5\)[/tex] is:
[tex]\[
x = 1, \quad x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i, \quad x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i
\][/tex]
