Let α, β, γ be the real roots of the equation, x³ + ax² + bx + c = 0, (a, b, c ∈ R and a, b ≠ 0). If

Q: Wait, why does a non-trivial solution imply the determinant is zero?

Think of the system as $M X = 0$. If $\det(M) \neq 0$, the matrix is invertible, meaning $X = M^{-1} 0 = 0$, which is the trivial solution. To escape the trivial, we need the matrix to be singular, so $\det(M) = 0$.

Q: Is there a faster way to expand that determinant?

Yes, use the property of circulant determinants. The value is $-(\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma)$. Memorizing this identity saves precious minutes in the exam hall.

Q: Why do we reject the case where $a=0$?

The problem explicitly states $a \neq 0$. Always check your constraints; they are the guardrails of your solution. If $a=0$, the equation $a(a^2 - 3b) = 0$ would be satisfied regardless of $b$, but the problem forces us into the specific case where $a^2 - 3b = 0$.

Question

Let $\alpha, \beta, \gamma$ be the real roots of the equation, $x^3 + ax^2 + bx + c = 0, (a, b, c \in \mathbb{R} 	ext{ and } a, b 
eq 0)$. If the system of equations (in $u, v, w$) given by $\alpha u + \beta v + \gamma w = 0, \beta u + \gamma v + \alpha w = 0, \gamma u + \alpha v + \beta w = 0$ has non-trivial solution, then the value of $\frac{a^2}{b}$ is

Options:

(A) 5

(B) 3

(C) 1

(D) 0

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Polynomials

Welcome, future engineer. Today, we are not just solving a problem; we are uncovering the hidden symmetry of polynomials. Imagine you are standing at the intersection of Algebra and Linear Algebra.

You have a cubic equation, $x^3 + ax^2 + bx + c = 0$, with roots $\alpha, \beta, \gamma$. Then, you are presented with a system of linear equations that seems to dance around these roots. This is the beauty of JEE Advanced—it forces you to connect disparate concepts.

### The Homogeneous Mystery

We start with a system of equations in $u, v, w$:
$$\alpha u + \beta v + \gamma w = 0$$
$$\beta u + \gamma v + \alpha w = 0$$
$$\gamma u + \alpha v + \beta w = 0$$

Because the right-hand side is zero, this is a homogeneous system. In the world of linear algebra, a homogeneous system $Ax = 0$ always has the trivial solution $u=v=w=0$.

But the problem demands a **non-trivial solution**. For a non-trivial solution to exist, the coefficient matrix must be singular. In other words, its determinant $D$ must be zero:

$$D = \begin{vmatrix} \alpha & \beta & \gamma \ \beta & \gamma & \alpha \ \gamma & \alpha & \beta \end{vmatrix} = 0$$

### The Circulant Shortcut

This determinant is a special beast known as a circulant determinant. The expansion of this specific determinant is a well-known result:

$$D = -(\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma) = 0$$

Since $D=0$, we have $\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = 0$. Now, recall the algebraic identity:

$$\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - (\alpha\beta + \beta\gamma + \gamma\alpha))$$

This factorization is the key that unlocks the door.

### The Vieta Bridge

Now, we bridge the gap back to our cubic equation. Vieta's formulas are our best friends here. For $x^3 + ax^2 + bx + c = 0$, we know that:

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

We also need the sum of squares, $\alpha^2 + \beta^2 + \gamma^2$. Using the identity $(\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha)$, we substitute our Vieta values:

$$(-a)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2b$$
$$\alpha^2 + \beta^2 + \gamma^2 = a^2 - 2b$$

### The Final Simplification

We are almost there. Substitute these expressions back into our factored determinant equation:

$$(-a) [ (a^2 - 2b) - (b) ] = 0$$

This simplifies beautifully to $-a(a^2 - 3b) = 0$. Since the problem explicitly states $a 
eq 0$, we can safely divide by $-a$.

This leaves us with $a^2 - 3b = 0$, or $a^2 = 3b$. Finally, since $b 
eq 0$, we divide by $b$ to find the ratio:

$$\frac{a^2}{b} = 3$$

It is elegant, it is precise, and it is the kind of problem that reminds us why we love mathematics. Keep practicing, and keep looking for those hidden symmetries. The final answer is **3**.

Select Answer:

Visualized Solution

The Cubic Equation

The Homogeneous System

Condition for Non-Trivial Solution

Setting up the Determinant

Expanding the Circulant Determinant

Factoring the Determinant

Applying Vieta's Formulas

Calculating Sum of Squares

Substituting into the Determinant Equation

Simplifying the Equation

Finding the Final Ratio $\frac{a ^{2}}{b}$

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

The Hidden Symmetry of Polynomials

The Homogeneous Mystery

The Circulant Shortcut

The Vieta Bridge

The Final Simplification

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