Detailed Solution

This problem beautifully combines concepts from the theory of equations (specifically, Vieta's formulas for polynomial roots) and linear algebra (conditions for non-trivial solutions of a system of homogeneous linear equations).

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1. Key Concepts and Formulas

Before diving into the solution, let's recall the fundamental principles we'll be using:

• Vieta's Formulas for a Cubic Equation: If $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3 + Ax^2 + Bx + C = 0$, then:

  • Sum of roots: $\alpha + \beta + \gamma = -A$
  • Sum of products of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = B$
  • Product of roots: $\alpha\beta\gamma = -C$

• Non-Trivial Solutions for Homogeneous Linear Equations: A system of homogeneous linear equations of the form $Ax = 0$ (where $A$ is a square matrix and $x$ is a column vector of variables) has non-trivial solutions (i.e., solutions other than $x=0$) if and only if the determinant of the coefficient matrix $A$ is zero, i.e., $\det(A) = 0$.

• Algebraic Identity: A crucial identity for the sum of cubes is: $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ This identity can also be written using the expansion of a sum squared: $x^2+y^2+z^2-xy-yz-zx = (x+y+z)^2 - 3(xy+yz+zx)$ Combining these, we get: $x^3 + y^3 + z^3 - 3xyz = (x+y+z)[(x+y+z)^2 - 3(xy+yz+zx)]$

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2. Step-by-Step Solution

Step 1: Apply Vieta's Formulas to the Given Cubic Equation

The given cubic equation is $x^3 + ax^2 + bx + c = 0$, and its roots are $\alpha, \beta, \gamma$. Comparing this with the general form $x^3 + Ax^2 + Bx + C = 0$, we have $A=a$, $B=b$, and $C=c$.

Using Vieta's formulas:

• Sum of the roots: $\alpha + \beta + \gamma = -a$
• Sum of products of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = b$
• Product of the roots: $\alpha\beta\gamma = -c$

Step 2: Set up the Condition for Non-Trivial Solutions

We are given a system of three homogeneous linear equations in $u, v, w$:

1. $\alpha u + \beta v + \gamma w = 0$
2. $\beta u + \gamma v + \alpha w = 0$
3. $\gamma u + \alpha v + \beta w = 0$

For this system to have non-trivial solutions (i.e., solutions where $u, v, w$ are not all zero), the determinant of its coefficient matrix must be zero.

The coefficient matrix is: $M = \begin{pmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{pmatrix}$

The condition for non-trivial solutions is $\det(M) = 0$. $\left| \begin{matrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{matrix} \right| = 0$

Step 3: Evaluate the Determinant

We expand the $3 \times 3$ determinant along the first row: $\alpha \left| \begin{matrix} \gamma & \alpha \\ \alpha & \beta \end{matrix} \right| - \beta \left| \begin{matrix} \beta & \alpha \\ \gamma & \beta \end{matrix} \right| + \gamma \left| \begin{matrix} \beta & \gamma \\ \gamma & \alpha \end{matrix} \right| = 0$

Now, we evaluate the $2 \times 2$ determinants: $\alpha(\gamma \cdot \beta - \alpha \cdot \alpha) - \beta(\beta \cdot \beta - \alpha \cdot \gamma) + \gamma(\beta \cdot \alpha - \gamma \cdot \gamma) = 0$ $\alpha(\beta\gamma - \alpha^2) - \beta(\beta^2 - \alpha\gamma) + \gamma(\alpha\beta - \gamma^2) = 0$

Distribute the terms: $\alpha\beta\gamma - \alpha^3 - \beta^3 + \alpha\beta\gamma + \alpha\beta\gamma - \gamma^3 = 0$

Combine like terms: $3\alpha\beta\gamma - (\alpha^3 + \beta^3 + \gamma^3) = 0$

Rearranging this equation, we get the crucial relationship: $\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = 0$

Step 4: Apply the Algebraic Identity

We now use the algebraic identity for the sum of cubes: $\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)[(\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha)]$

Since we found that $\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = 0$, we can substitute this into the identity: $(\alpha+\beta+\gamma)[(\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha)] = 0$

Step 5: Substitute Vieta's Formulas and Solve for $\frac{a^2}{b}$

Now, we substitute the expressions from Vieta's formulas (from Step 1) into the equation from Step 4:

• $\alpha + \beta + \gamma = -a$
• $\alpha\beta + \beta\gamma + \gamma\alpha = b$

Substituting these into the identity: $(-a)[(-a)^2 - 3(b)] = 0$ $(-a)[a^2 - 3b] = 0$

This equation implies two possibilities:

1. $-a = 0 \implies a = 0$
2. $a^2 - 3b = 0 \implies a^2 = 3b$

The problem statement explicitly mentions that $a \ne 0$. Therefore, we must discard the first possibility ($a=0$).

This leaves us with the second possibility: $a^2 = 3b$

The question asks for the value of $\frac{a^2}{b}$. Since $b \ne 0$ (otherwise $a^2=0 \Rightarrow a=0$, which contradicts $a \ne 0$), we can divide by $b$: $\frac{a^2}{b} = 3$

Thus, the value of $\frac{a^2}{b}$ is 3.

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3. Tips for Success and Common Pitfalls

• Master Vieta's Formulas: These are fundamental for problems relating roots and coefficients of polynomials.
• Understand Determinant Conditions: Remember that $\det(A)=0$ is the key for non-trivial solutions in homogeneous systems. For non-homogeneous systems, Cramer's rule or matrix inversion are relevant.
• Know Algebraic Identities: The identity $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ is extremely common in JEE problems. Be comfortable with its alternative forms.
• Careful with Expansion: When expanding determinants, especially $3 \times 3$, be meticulous with signs and terms to avoid errors.
• Utilize All Given Conditions: The condition $a \ne 0$ was crucial to select the correct branch of the solution. Always check if all constraints have been applied.
• Recognize Patterns: The determinant in this problem is a special type (a circulant determinant) whose value is often related to sums of powers of roots. Recognizing such patterns can sometimes speed up the evaluation, but direct expansion is always reliable.

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4. Summary and Key Takeaway

This problem is an excellent example of how different areas of mathematics (algebra of polynomials and linear algebra) are interconnected. The core idea was to:

1. Translate the properties of the polynomial roots into relationships between its coefficients using Vieta's formulas.
2. Translate the condition for non-trivial solutions of the linear system into a condition on the determinant of its coefficient matrix.
3. Evaluate this determinant, which led to a standard algebraic identity.
4. Substitute the Vieta's formulas back into the identity to establish a relationship between the coefficients $a$ and $b$, ultimately leading to the desired value.

The solution highlights the importance of a strong foundation in both polynomial theory and matrix determinants, along with algebraic manipulation skills.

The final answer is $\boxed{3}$.