To prove the equation:

\[
\frac{1 + \tan^3 \theta}{1 + \tan \theta} = 1 - \tan \theta + \tan^2 \theta
\]

Let's follow the steps to verify whether the equation holds.

### Step 1: Simplify the Left-Hand Side (LHS)
We start by simplifying the left-hand side (LHS) of the equation:

\[
\text{LHS} = \frac{1 + \tan^3 \theta}{1 + \tan \theta}
\]

We can factor the numerator as:

\[
1 + \tan^3 \theta = (1 + \tan \theta)(1 - \tan \theta + \tan^2 \theta)
\]

This is a well-known factorization identity for cubic expressions, \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), with \( a = 1 \) and \( b = \tan \theta \).

Now the LHS becomes:

\[
\text{LHS} = \frac{(1 + \tan \theta)(1 - \tan \theta + \tan^2 \theta)}{1 + \tan \theta}
\]

### Step 2: Cancel common terms
Since \( 1 + \tan \theta \) appears both in the numerator and denominator, we can cancel it out (assuming \( 1 + \tan \theta \neq 0 \)):

\[
\text{LHS} = 1 - \tan \theta + \tan^2 \theta
\]

### Step 3: Compare with the Right-Hand Side (RHS)
The right-hand side (RHS) of the original equation is:

\[
\text{RHS} = 1 - \tan \theta + \tan^2 \theta
\]

Thus, we have:

\[
\text{LHS} = \text{RHS}
\]

### Conclusion
The equation is proven to be true.

\[
\boxed{\frac{1 + \tan^3 \theta}{1 + \tan \theta} = 1 - \tan \theta + \tan^2 \theta}
\]

Let me know if you want more details or if you have any questions.

Here are 5 related questions for further practice:
1. Prove the identity: \( \frac{1 - \tan^3 \theta}{1 - \tan \theta} = 1 + \tan \theta + \tan^2 \theta \).
2. Simplify: \( \frac{1 + \cot^3 \theta}{1 + \cot \theta} \).
3. Verify: \( \frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} = 1 - \sin \theta \cos \theta \).
4. Solve \( \tan^2 \theta + \tan \theta = 0 \) for \( \theta \).
5. Prove that \( 1 + \tan^2 \theta = \sec^2 \theta \).

**Tip**: When proving trigonometric identities, factorizing expressions and using known identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) are very helpful strategies.

Discover how to prove the trigonometric identity (1 + tan^3 θ) / (1 + tan θ) = 1 - tan θ + tan^2 θ with a detailed step-by-step solution. This problem involves key trigonometric concepts, including factorization and tangent identities. Suitable for high school students (Grades 10-12).

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