To solve the cubic equation [tex]\( x^3 - (3 + \sqrt{3}) x + 3 = 0 \)[/tex], we aim to find the values of [tex]\( x \)[/tex] that satisfy it.
Let's denote the cubic equation as:
[tex]\[ f(x) = x^3 - (3 + \sqrt{3}) x + 3 \][/tex]
To solve this equation, we generally apply algebraic methods or numerical techniques. The solutions to a cubic equation can be complex (involving imaginary numbers) or real.
For this specific cubic equation, the solutions are complex expressions. Here are the solutions determined using algebraic techniques (e.g., Cardano's formula) and simplification:
1. First solution:
[tex]\[
x_1 = -\frac{-\frac{1}{2} + \frac{\sqrt{3}i}{2}}{3} \left(\frac{81}{2} + \frac{\sqrt{6561 - 4 \cdot (3\sqrt{3} + 9)^3}}{2}\right)^{\frac{1}{3}} - \frac{3 \sqrt{3} + 9}{3 \left( -\frac{1}{2} + \frac{\sqrt{3}i}{2} \right)\left(\frac{81}{2} + \frac{\sqrt{6561 - 4 \cdot (3\sqrt{3} + 9)^3}}{2}\right)^{\frac{1}{3}}}
\][/tex]
2. Second solution:
[tex]\[
x_2 = -\frac{3 \sqrt{3} + 9}{3 \left( -\frac{1}{2} - \frac{\sqrt{3}i}{2} \right)\left(\frac{81}{2} + \frac{\sqrt{6561 - 4 \cdot (3\sqrt{3} + 9)^3}}{2}\right)^{\frac{1}{3}}} - \frac{-\frac{1}{2} - \frac{\sqrt{3}i}{2}}{3} \left(\frac{81}{2} + \frac{\sqrt{6561 - 4 \cdot (3\sqrt{3} + 9)^3}}{2}\right)^{\frac{1}{3}}
\][/tex]
3. Third solution:
[tex]\[
x_3 = -\left( \frac{81}{2} + \frac{\sqrt{6561 - 4 \cdot (3\sqrt{3} + 9)^3}}{2} \right)^{\frac{1}{3}} \frac{1}{3} - \frac{3 \sqrt{3} + 9}{3 \left( \frac{81}{2} + \frac{\sqrt{6561 - 4 \cdot (3\sqrt{3} + 9)^3}}{2} \right)^{\frac{1}{3}}}
\][/tex]
These solutions involve intricate expressions with cube roots and complex numbers. Solving them manually involves using advanced algebraic techniques for cubics and handling complex arithmetic, but the steps involve breaking down these expressions using rules of algebra and properties of complex numbers.
We have determined three solutions ([tex]\( x_1, x_2, x_3 \)[/tex]) for the given cubic equation [tex]\( x^3 - (3 + \sqrt{3}) x + 3 = 0 \)[/tex]. Each of these solutions retains a complex form due to the nature of cube roots and the involvement of complex terms.
