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Step-by-Step Solution
Step 1: Identify the Roots and Their Known Relationships
Let the roots of the cubic equation
$x^3 + a x^2 + b x + c = 0$
be $ \alpha, \beta, \gamma$. Then from Vieta's formulas, we know:
$\alpha + \beta + \gamma = -a$
$\alpha \beta + \beta \gamma + \gamma \alpha = b$
$\alpha \beta \gamma = -c$
Step 2: Form the Coefficient Matrix for the Given System
The system of linear equations in $u, v, w$ is:
$$
\begin{cases}
\alpha u + \beta v + \gamma w = 0, \\
\beta u + \gamma v + \alpha w = 0, \\
\gamma u + \alpha v + \beta w = 0.
\end{cases}
$$
We represent this system in matrix form as $M \cdot \begin{pmatrix} u \\ v \\ w \end{pmatrix} = 0$, where
$$
M = \begin{pmatrix}
\alpha & \beta & \gamma \\
\beta & \gamma & \alpha \\
\gamma & \alpha & \beta
\end{pmatrix}.
$$
Step 3: Condition for Non-trivial Solutions
A non-trivial solution $\bigl(u, v, w\bigr) \ne (0,0,0)$ exists if and only if the determinant of $M$ is zero:
$$
\det(M) =
\begin{vmatrix}
\alpha & \beta & \gamma \\
\beta & \gamma & \alpha \\
\gamma & \alpha & \beta
\end{vmatrix}
= 0.
$$
Step 4: Evaluate the Determinant
It is a known result (or can be expanded) that:
$$
\alpha^3 + \beta^3 + \gamma^3 - 3\,\alpha\,\beta\,\gamma = 0.
$$
Alternatively, one can directly compute the determinant by expansion, but this identity is a standard shortcut for this cyclic form.
Step 5: Express in Terms of $a, b, c$
Recall the factorization:
$$
\alpha^3 + \beta^3 + \gamma^3 - 3\alpha \beta \gamma =
(\alpha + \beta + \gamma)\bigl[ (\alpha + \beta + \gamma)^2 - 3(\alpha \beta + \beta \gamma + \gamma \alpha)\bigr].
$$
Substituting Vieta's relations:
$$
(\alpha + \beta + \gamma) = -a,
$$
$$
(\alpha \beta + \beta \gamma + \gamma \alpha) = b,
$$
$$
\alpha \beta \gamma = -c,
$$
we get
$$
(\,-a)\Bigl[(-a)^2 - 3 b\Bigr] = 0.
$$
Step 6: Solve for $a^2 / b$
Since $a \ne 0$ (given), the factor $(-a)$ cannot be zero, so we must have
$$
(-a)^2 - 3b = 0
\quad\Longrightarrow\quad
a^2 = 3b
\quad\Longrightarrow\quad
\frac{a^2}{b} = 3.
$$
Answer: 3
