To find the expansion of $(e^{2it} - e^{-2it})^3$, let's proceed step by step:

Step 1: Let $x = e^{2it}$.

Thus, $e^{-2it} = \frac{1}{x}$.

Now the expression becomes: $(e^{2it} - e^{-2it})^3 = (x - \frac{1}{x})^3.$

Step 2: Simplify $(x - \frac{1}{x})^3$.

We will use the binomial expansion for cubes: $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.$ Here, $a = x$ and $b = \frac{1}{x}$, so we substitute these into the formula: $(x - \frac{1}{x})^3 = x^3 - 3x^2 \cdot \frac{1}{x} + 3x \cdot \frac{1}{x^2} - \frac{1}{x^3}.$

Step 3: Simplify each term.

$x^3 - 3x + 3\frac{1}{x} - \frac{1}{x^3}.$

Thus, the expanded form of $(e^{2it} - e^{-2it})^3$ is: $x^3 - 3x + 3\frac{1}{x} - \frac{1}{x^3}.$

Step 4: Substitute back $x = e^{2it}$.

Now we substitute $x = e^{2it}$ back into the expression: $e^{6it} - 3e^{2it} + 3e^{-2it} - e^{-6it}.$

So, the expansion of $(e^{2it} - e^{-2it})^3$ is: $e^{6it} - 3e^{2it} + 3e^{-2it} - e^{-6it}.$

Would you like further details on the steps or how this relates to trigonometric identities?

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Here are five related questions to explore:

1. How does the binomial expansion work for higher powers?
2. What are Euler's formulas for expressing $e^{it}$ in terms of sine and cosine?
3. How can you use this expansion in solving differential equations?
4. What is the significance of complex exponentials in wave theory?
5. How would this expansion change if you were working in polar coordinates?

Tip: Remember that complex exponentials $e^{ix}$ are closely related to trigonometric functions through Euler’s formula $e^{ix} = \cos x + i \sin x$.