The problem asks us to find the value of the product xyz given a determinant equation involving distinct real numbers x, y, and z.

Understanding the Determinant Equation

We are given the equation:

\[ \left| {\begin{array}{} x&{{x^2}}&{2 + {x^3}}\\ y&{{y^2}}&{2 + {y^3}}\\ z&{{z^2}}&{2 + {z^3}} \end{array}} \right| = 0 \]

The key to solving this is to use the properties of determinants, specifically the linearity property which allows us to split a determinant if one column is a sum of terms.

Determinant Expansion Using Properties

We can split the determinant based on the third column, which is a sum of '2' and '{term}^3'. Let the determinant be denoted by D.

Using the property $\det(C_1, C_2, C_3 + C'_3) = \det(C_1, C_2, C_3) + \det(C_1, C_2, C'_3)$, we can rewrite D as the sum of two determinants:

$$ D = \left| {\begin{array}{} x&{{x^2}}&{2}\\ y&{{y^2}}&{2}\\ z&{{z^2}}&{2} \end{array}} \right| + \left| {\begin{array}{} x&{{x^2}}&{{x^3}}\\ y&{{y^2}}&{{y^3}}\\ z&{{z^2}}&{{z^3}} \end{array}} \right| $$

Let's evaluate each determinant separately.

Evaluating the First Determinant

Let the first determinant be D1:

$$ D_1 = \left| {\begin{array}{} x&{{x^2}}&{2}\\ y&{{y^2}}&{2}\\ z&{{z^2}}&{2} \end{array}} \right| $$

We can factor out the common constant '$2$' from the third column:

$$ D_1 = 2 \left| {\begin{array}{} x&{{x^2}}&{1}\\ y&{{y^2}}&{1}\\ z&{{z^2}}&{1} \end{array}} \right| $$

This determinant is related to the Vandermonde determinant. The standard Vandermonde determinant is:

\[ \left| {\begin{array}{} 1&{a}&{{a^2}}\\ 1&{{b}}&{{b^2}}\\ 1&{{c}}&{{c^2}} \end{array}} \right| = (b-a)(c-a)(c-b) \]

By swapping columns, we can see that:

\[ \left| {\begin{array}{} x&{{x^2}}&{1}\\ y&{{y^2}}&{1}\\ z&{{z^2}}&{1} \end{array}} \right| = \left| {\begin{array}{} 1&{x}&{{x^2}}\\ 1&{{y}}&{{y^2}}\\ 1&{{z}}&{{z^2}} \end{array}} \right| = (y-x)(z-x)(z-y) \]

So, the first determinant evaluates to:

$$ D_1 = 2 (y-x)(z-x)(z-y) $$

Evaluating the Second Determinant

Let the second determinant be D2:

$$ D_2 = \left| {\begin{array}{} x&{{x^2}}&{{x^3}}\\ y&{{y^2}}&{{y^3}}\\ z&{{z^2}}&{{z^3}} \end{array}} \right| $$

We can factor out '$x$' from the first row, '$y$' from the second row, and '$z$' from the third row:

$$ D_2 = x \left| {\begin{array}{} 1 & {{x^2}} & {{x^3}} \\ y & {{y^2}} & {{y^3}} \\ z & {{z^2}} & {{z^3}} \end{array}} \right| $$ $$ D_2 = xy \left| {\begin{array}{} 1 & 1 & {{x^3}} \\ 1 & y & {{y^3}} \\ 1 & z & {{z^3}} \end{array}} \right| $$ $$ D_2 = xyz \left| {\begin{array}{} 1 & 1 & 1 \\ 1 & y & {{y^2}} \\ 1 & z & {{z^2}} \end{array}} \right| $$

This calculation seems incorrect. Let's factor differently. Factor '$x$' from Column 1, '$x$' from Column 2, '$x$' from Column 3? No.

Let's factor out $x$ from Row 1, $y$ from Row 2, and $z$ from Row 3 directly:

$$ D_2 = x \cdot y \cdot z \left| {\begin{array}{} 1 & x & {{x^2}} \\ 1 & y & {{y^2}} \\ 1 & z & {{z^2}} \end{array}} \right| $$

This is $xyz$ multiplied by the Vandermonde determinant we saw earlier.

$$ D_2 = xyz (y-x)(z-x)(z-y) $$

Combining and Solving for xyz

Now we substitute the evaluated determinants back into the original equation $D = D_1 + D_2 = 0$:

$$ 2 (y-x)(z-x)(z-y) + xyz (y-x)(z-x)(z-y) = 0 $$

We can factor out the common term $(y-x)(z-x)(z-y)$:

$$ (y-x)(z-x)(z-y) [2 + xyz] = 0 $$

The problem states that x, y, and z are distinct real numbers. This means that:

• $y - x \neq 0$
• $z - x \neq 0$
• $z - y \neq 0$

Since the product of these three non-zero terms is non-zero, for the entire equation to be zero, the remaining factor must be zero:

$$ 2 + xyz = 0 $$

Solving for xyz, we get:

$$ xyz = -2 $$