If \( \alpha, \beta, \gamma \) are the roots of the equation \( x^3 - 3x + 1 = 0 \), then \( \frac{\alpha}{1-\alpha} + \frac{\beta}{1-\beta} + \frac{\gamma}{1-\gamma} = \)
Show Hint
Transformation of equations is the most robust technique for finding sums of rational functions of roots. It avoids complex algebra and symmetric sum calculations.
Step 1: Choose the transformation method.
Instead of grinding symmetric sums, we build a new cubic whose roots are $y = \dfrac{x}{1-x}$. Then the answer is simply the sum of those new roots. Step 2: Invert the relation.
From $y = \dfrac{x}{1-x}$, solve for $x$: $y(1-x) = x \Rightarrow y = x(1+y) \Rightarrow x = \dfrac{y}{1+y}$. Step 3: Substitute into the original cubic.
Put $x = \dfrac{y}{1+y}$ into $x^3 - 3x + 1 = 0$ and multiply through by $(1+y)^3$: \[ y^3 - 3y(1+y)^2 + (1+y)^3 = 0. \] Step 4: Expand carefully.
$-3y(1+2y+y^2) = -3y - 6y^2 - 3y^3$ and $(1+y)^3 = 1 + 3y + 3y^2 + y^3$. Adding $y^3$ from the first term: \[ (y^3 - 3y^3 + y^3) + (-6y^2 + 3y^2) + (-3y + 3y) + 1 = -y^3 - 3y^2 + 1 = 0. \] Step 5: Tidy into standard form.
Multiply by $-1$: $y^3 + 3y^2 - 1 = 0$. Step 6: Read off the sum of roots.
For $y^3 + 3y^2 + 0\cdot y - 1 = 0$, the sum of roots is $-\dfrac{3}{1} = -3$. That is exactly $\sum \dfrac{\alpha}{1-\alpha}$, option (C).
\[ \boxed{-3} \]