Count Real Roots of Given Determinant Equation

To find the total count of distinct real numbers $x \in \mathbb{R}$ satisfying the given determinant equation:

$$\left| \begin{array}{ccc} x & x^2 & 1 + x^3 \\ 2x & 4x^2 & 1 + 8x^3 \\ 3x & 9x^2 & 1 + 27x^3 \end{array} \right| = 10$$

we can split the determinant into the sum of two determinants using the linearity property along the third column:

$$\Delta = \left| \begin{array}{ccc} x & x^2 & 1 \\ 2x & 4x^2 & 1 \\ 3x & 9x^2 & 1 \end{array} \right| + \left| \begin{array}{ccc} x & x^2 & x^3 \\ 2x & 4x^2 & 8x^3 \\ 3x & 9x^2 & 27x^3 \end{array} \right|$$

Let us evaluate each determinant separately:

1. First Determinant ($\Delta_1$): Factoring out $x$ from the first column $C_1$ and $x^2$ from the second column $C_2$:

$$\Delta_1 = x \cdot x^2 \left| \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 3 & 9 & 1 \end{array} \right| = x^3 \left| \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 3 & 9 & 1 \end{array} \right|$$

Evaluating the $3 \times 3$ matrix determinant:

$$\left| \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 3 & 9 & 1 \end{array} \right| = 1(4 - 9) - 1(2 - 3) + 1(18 - 12) = -5 + 1 + 6 = 2$$

Thus, $\Delta_1 = 2x^3$

1. Second Determinant ($\Delta_2$): Factoring out $x$ from $R_1$, $2x$ from $R_2$, and $3x$ from $R_3$:

$$\Delta_2 = (x)(2x)(3x) \left| \begin{array}{ccc} 1 & x & x^2 \\ 1 & 2x & 4x^2 \\ 1 & 3x & 9x^2 \end{array} \right| = 6x^3 \cdot x^3 \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \end{array} \right|$$

The determinant is a standard Vandermonde determinant:

$$\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \end{array} \right| = (2 - 1)(3 - 1)(3 - 2) = 1 \cdot 2 \cdot 1 = 2$$

Thus, $\Delta_2 = 6x^6 \cdot 2 = 12x^6$

Combining both parts, the original equation becomes: $\Delta_1 + \Delta_2 = 10$ $12x^6 + 2x^3 = 10$

Dividing both sides by $2$: $6x^6 + x^3 - 5 = 0$

Let $t = x^3$. The equation simplifies to a quadratic in $t$: $6t^2 + t - 5 = 0$ $(6t - 5)(t + 1) = 0$

This gives two possible values for $t$: $t = \frac{5}{6} \quad \text{or} \quad t = -1$

Now, substituting $x^3 = t$:

1. $x^3 = \frac{5}{6} \implies x = \left(\frac{5}{6}\right)^{1/3} \in \mathbb{R}$
2. $x^3 = -1 \implies x = -1 \in \mathbb{R}$

Both roots are distinct real numbers. Hence, the total count of distinct real numbers $x \in \mathbb{R}$ satisfying the given equation is $2$.