1. Key Concepts and Formulas

• Reciprocal Equations: A polynomial equation $P(x) = 0$ is reciprocal if its coefficients are symmetric (i.e., $a_i = a_{n-i}$). If $r$ is a root, then $1/r$ is also a root.
• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.
• Sum of Cubes Identity: $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$.

2. Step-by-Step Solution

Step 1: Recognize and exploit the reciprocal nature of the equation. The given equation is $x^4 - 3x^3 - 2x^2 + 3x + 1 = 0$. The coefficients are symmetric (1, -3, -2, 3, 1), indicating a reciprocal equation. This suggests dividing by $x^2$ to simplify. Since $x=0$ is not a root, we can safely divide by $x^2$.

$\frac{x^4 - 3x^3 - 2x^2 + 3x + 1}{x^2} = 0$ $x^2 - 3x - 2 + \frac{3}{x} + \frac{1}{x^2} = 0$

Step 2: Rearrange and substitute to form a quadratic equation. Group terms with similar powers of $x$: $(x^2 + \frac{1}{x^2}) - 3(x - \frac{1}{x}) - 2 = 0$ Let $y = x - \frac{1}{x}$. Then $y^2 = x^2 - 2 + \frac{1}{x^2}$, so $x^2 + \frac{1}{x^2} = y^2 + 2$. Substitute these into the equation: $(y^2 + 2) - 3y - 2 = 0$ $y^2 - 3y = 0$

Step 3: Solve the quadratic equation for y. Factor the quadratic equation: $y(y - 3) = 0$ Thus, $y = 0$ or $y = 3$.

Step 4: Solve for x using the values of y. Case 1: $y = 0$ $x - \frac{1}{x} = 0$ $x^2 - 1 = 0$ $x^2 = 1$ $x = \pm 1$

Case 2: $y = 3$ $x - \frac{1}{x} = 3$ $x^2 - 1 = 3x$ $x^2 - 3x - 1 = 0$ Let the roots of this quadratic be $\alpha$ and $\beta$.

Step 5: Apply Vieta's formulas and the sum of cubes identity. We have the four roots: $1, -1, \alpha, \beta$. We want to find $1^3 + (-1)^3 + \alpha^3 + \beta^3$. Since $1^3 = 1$ and $(-1)^3 = -1$, $1^3 + (-1)^3 = 0$. So, we only need to find $\alpha^3 + \beta^3$.

For the quadratic $x^2 - 3x - 1 = 0$, by Vieta's formulas: $\alpha + \beta = 3$ and $\alpha\beta = -1$. Using the sum of cubes identity: $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = (3)^3 - 3(-1)(3) = 27 + 9 = 36$.

Step 6: Calculate the total sum of cubes. The sum of the cubes of all roots is $1^3 + (-1)^3 + \alpha^3 + \beta^3 = 0 + 36 = 36$.

3. Common Mistakes & Tips

• Forgetting roots: Make sure to account for all four roots of the quartic equation.
• Sign errors in Vieta's formulas: Double-check the signs when applying Vieta's formulas.
• Incorrectly applying algebraic identities: Ensure you use the correct identity for the sum of cubes.

4. Summary

By recognizing the reciprocal nature of the equation, we simplified it into a quadratic. Solving for $x$ yielded two roots directly, while the other two were expressed as roots of another quadratic. Applying Vieta's formulas and the sum of cubes identity allowed us to efficiently compute the sum of the cubes of all roots without explicitly solving for them.

The final answer is $\boxed{36}$.