The total number of distinct x in R for which |[x, x^2, 1+x^3], [2x, 4x^2, 1+8x^3], [3x, 9x^2, 1+27x

Q: Wait, why can't I just expand the determinant directly? It's only a $3 \times 3$ matrix, right?

Technically, you could, but you would be walking into a trap. Expanding it directly would lead to a massive algebraic expression involving high-degree polynomials that would be incredibly easy to mess up. In JEE Advanced, the goal isn't just to solve the problem; it's to solve it efficiently. Using determinant properties is the 'pro' way—it reduces the complexity from a nightmare of algebra to a simple quadratic equation.

Q: Why do we get $x^3$ and $x^6$ outside the determinants? I'm confused about the factoring.

Think of it this way: when you factor a constant $k$ out of a row or column in a determinant, you are factoring it out of the *entire* determinant. In the first determinant, we pulled $x$ from the first column and $x^2$ from the second column. Since $x \cdot x^2 = x^3$, that $x^3$ comes out as a multiplier. In the second determinant, we pulled $x$ from $C_1$, $x^2$ from $C_2$, and $x^3$ from $C_3$. Multiplying those together gives $x^6$. It's just basic algebra applied to the structure of the matrix!

Q: How do I know if $x^3 = 5/6$ gives a real root? Do I need to worry about imaginary numbers?

That is a fantastic question! Unlike square roots, where a negative number inside the root leads to imaginary results, cube roots are perfectly happy with any real number. If $x^3 = k$, there is always exactly one real solution for $x$, which is $x = \sqrt[3]{k}$. Since $5/6$ is a real number and $-1$ is a real number, both cases yield valid, distinct real roots. No imaginary numbers here!

Question

The total number of distinct $x \in \mathbb{R}$ for which $\begin{vmatrix} x & x^2 & 1+x^3 \ 2x & 4x^2 & 1+8x^3 \ 3x & 9x^2 & 1+27x^3 \end{vmatrix} = 10$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of variables and powers. You see a $3 	imes 3$ determinant, and your instinct might be to start expanding it row by row.

Stop! Take a breath. In JEE Advanced, whenever you see a determinant that looks overly complicated, it is almost never meant to be expanded directly. There is a hidden structure, a geometric elegance waiting to be revealed.

Look at the given equation:

$$\begin{vmatrix} x & x^2 & 1+x^3 \ 2x & 4x^2 & 1+8x^3 \ 3x & 9x^2 & 1+27x^3 \end{vmatrix} = 10$$

Focus your gaze entirely on the third column. Every single element is a sum: $1 + (nx)^3$. This is not a coincidence; it is a breadcrumb trail left by the examiner.

In the language of linear algebra, this column is a sum of two vectors. This is the moment where we apply the **Linearity Property of Determinants**. We can split this single, intimidating determinant into the sum of two separate, much friendlier determinants.

### The Great Splitting

By keeping the first two columns identical, we can isolate the 'one's and the 'cube' terms. Imagine we are peeling back layers of an onion. We write:

$$\begin{vmatrix} x & x^2 & 1 \ 2x & 4x^2 & 1 \ 3x & 9x^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & x^3 \ 2x & 4x^2 & 8x^3 \ 3x & 9x^2 & 27x^3 \end{vmatrix} = 10$$

Suddenly, the problem feels lighter, doesn't it? We have transformed one complex beast into two smaller, manageable creatures.

### The Power of Factoring

Now, let us look at these new determinants. In the first one, look at the first column: $x, 2x, 3x$. We can factor out an $x$.

Look at the second column: $x^2, 4x^2, 9x^2$. We can factor out an $x^2$. When we pull these out, they multiply to become $x^3$ outside the determinant. We are left with:

$$x^3 \begin{vmatrix} 1 & 1 & 1 \ 2 & 4 & 1 \ 3 & 9 & 1 \end{vmatrix}$$

We do the exact same thing for the second determinant. Factor $x$ from $C_1$, $x^2$ from $C_2$, and $x^3$ from $C_3$. This gives us $x^6$ outside! We are left with:

$$x^6 \begin{vmatrix} 1 & 1 & 1 \ 2 & 4 & 8 \ 3 & 9 & 27 \end{vmatrix}$$

### The Numerical Victory

Now, the variables are safely outside. We are left with pure arithmetic. Evaluating the first determinant:

$$D_1 = 1(4-9) - 1(2-3) + 1(18-12) = -5 + 1 + 6 = 2$$

And the second determinant:

$$D_2 = 1(108-72) - 1(54-24) + 1(18-12) = 36 - 30 + 6 = 12$$

Our equation has collapsed from a matrix nightmare into a beautiful, simple polynomial: $2x^3 + 12x^6 = 10$. Dividing by $2$, we get $6x^6 + x^3 - 5 = 0$.

### The Quadratic Disguise

This is the final act. We have a degree-six polynomial, but notice the powers: $6$ and $3$. The higher power is exactly double the lower one.

This is a quadratic in disguise! Let $t = x^3$. Our equation becomes $6t^2 + t - 5 = 0$.

Factoring this, we get $(6t-5)(t+1) = 0$. This gives us $t = 5/6$ and $t = -1$.

Since $t = x^3$, we have $x^3 = 5/6$ and $x^3 = -1$. Both equations yield exactly one real root each.

Thus, we have found **2** distinct real values for $x$. You see? With the right tools and a calm mind, even the most intimidating problems surrender. Keep practicing, keep visualizing, and keep falling in love with the logic!

Animated Solution for Mathematics - Matrices and Determinants: The total number of distinct $x \in R$ for which $x 2 x 3 x x^{2} 4 x^{2} 9 x^{2} 1 + x^{3} 1 + 8 x^{3} 1 + 27 x^{3} = 10$ is

Enter Numerical Value:

Visualized Solution

Analyze the Determinant Structure

Apply the Splitting Property

Split the Determinant

Factor Out Common Terms (Part 1)

Factor Out Common Terms (Part 2)

Evaluate the First Determinant

Evaluate the Second Determinant

Form the Polynomial Equation

Substitute to Form a Quadratic

Solve the Quadratic Equation

Find the Values of $t$

Find Real Roots of $x$

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Great Splitting

The Power of Factoring

The Numerical Victory

The Quadratic Disguise

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