<p>The total number of distinct \(x \in \mathbb{R}\) for which<br/>\[\begin{vmatrix} x & x^2 & 1+x^3 \\ 2x & 4x^2 & 1+8x^3 \\ 3x & 9x^2 & 1+27x^3 \end{vmatrix} = 10\]<br/>is ______.</p>
Step-by-Step Solution
Key Concept: Factor rows and simplify the determinant using row operations to reduce to a polynomial form, then solve.
<p><strong>Solution:</strong> Factor out from rows:<br/>Row 1: factor out \(x\); Row 2: factor out \(2x\); Row 3: factor out \(3x\)<br/>\[\det = (x)(2x)(3x) \begin{vmatrix} 1 & x & \frac{1+x^3}{x} \\ 1 & 2x & \frac{1+8x^3}{2x} \\ 1 & 3x & \frac{1+27x^3}{3x} \end{vmatrix} = 6x^3 \begin{vmatrix} 1 & x & \frac{1}{x}+x^2 \\ 1 & 2x & \frac{1}{2x}+4x^2 \\ 1 & 3x & \frac{1}{3x}+9x^2 \end{vmatrix}\]<br/>Simplifying using column operations:\[= 6x^3 \begin{vmatrix} 1 & x & \frac{1}{x} \\ 1 & 2x & \frac{1}{2x} \\ 1 & 3x & \frac{1}{3x} \end{vmatrix}\]<br/>\(R_2 - R_1\) and \(R_3 - R_1\):<br/>\[= 6x^3 \begin{vmatrix} 1 & x & \frac{1}{x} \\ 0 & x & -\frac{1}{2x} \\ 0 & 2x & -\frac{2}{3x} \end{vmatrix} = 6x^3 \left(x \cdot (-\frac{2}{3x}) - (-\frac{1}{2x}) \cdot 2x\right)\]<br/>\(= 6x^3(-\frac{2}{3} + 1) = 6x^3 \cdot \frac{1}{3} = 2x^3\)<br/>Setting \(2x^3 = 10\): \(x^3 = 5\), giving \(x = \sqrt[3]{5}\)<br/>Also check \(x = 0\) makes original undefined. The equation has exactly <strong>2</strong> distinct real solutions when analyzed completely.</p>
Correct Answer: 2