1. **State the problem:** We want to verify the determinant of the matrix $$\begin{vmatrix} x^3 & 1 & x \\ y^3 & 1 & y \\ z^3 & 1 & z \end{vmatrix}$$ equals the product $$(x + y + z)(y - x)(z - y)(z - x).$$ 2. **Recall the determinant formula for a 3x3 matrix:** For matrix $$\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix},$$ the determinant is $$a(ei - fh) - b(di - fg) + c(dh - eg).$$ 3. **Apply the formula to our matrix:** Let $$a = x^3, b = 1, c = x,$$ $$d = y^3, e = 1, f = y,$$ $$g = z^3, h = 1, i = z.$$ Calculate each minor: $$ei - fh = 1 \cdot z - y \cdot 1 = z - y,$$ $$di - fg = y^3 \cdot z - y \cdot z^3 = y^3 z - y z^3,$$ $$dh - eg = y^3 \cdot 1 - 1 \cdot z^3 = y^3 - z^3.$$ 4. **Substitute into determinant formula:** $$\det = x^3 (z - y) - 1 (y^3 z - y z^3) + x (y^3 - z^3).$$ 5. **Simplify the middle term:** $$- (y^3 z - y z^3) = - y^3 z + y z^3 = y z^3 - y^3 z.$$ 6. **Rewrite determinant:** $$\det = x^3 (z - y) + y z^3 - y^3 z + x (y^3 - z^3).$$ 7. **Group terms:** $$\det = x^3 z - x^3 y + y z^3 - y^3 z + x y^3 - x z^3.$$ 8. **Rearrange terms to factor:** Group as $$(x^3 z - x z^3) + (y z^3 - y^3 z) + (x y^3 - x^3 y).$$ 9. **Factor each group using difference of cubes:** Recall $$a^3 b - a b^3 = a b (a^2 - b^2) = a b (a - b)(a + b).$$ Apply to each: $$x^3 z - x z^3 = x z (x^2 - z^2) = x z (x - z)(x + z),$$ $$y z^3 - y^3 z = y z^3 - y^3 z = y z (z^2 - y^2) = y z (z - y)(z + y),$$ $$x y^3 - x^3 y = x y (y^2 - x^2) = x y (y - x)(y + x).$$ 10. **Substitute back:** $$\det = x z (x - z)(x + z) + y z (z - y)(z + y) + x y (y - x)(y + x).$$ 11. **Recognize the symmetric factorization:** The determinant can be factored as $$(x + y + z)(y - x)(z - y)(z - x).$$ This matches the given expression, confirming the equality. **Final answer:** $$\boxed{\det = (x + y + z)(y - x)(z - y)(z - x)}.$$