Number of positive integral solutions of the equation $\begin{vmatrix} x^3 + 1 & x^2y & x^2z \\ xy^2 & y^3 + 1 & y^2z \\ x^2 & yz^2 & z^3 + 1 \end{vmatrix} = 30$ are
Step-by-Step Solution
Key Concept: Column operations reduce the determinant to x³ + y³ + z³ + 1, which equals 30, giving x³ + y³ + z³ = 29. Finding positive integer solutions requires decomposing 29 as a sum of three perfect cubes (1³, 2³, 3³, etc.).
We evaluate the given $3×3$ determinant by performing column operations: $C_2 → C_2 - yC_1$ and $C_3 → C_3 - zC_1$. After simplification, the determinant reduces to $x^3 + y^3 + z^3 + 1 - y^3z^3$. Given the constraint $x^3 + y^3 + z^3 + 1 = 30$ and $x^3 + y^3 + z^3 = 29$, we substitute to find the determinant value equals $x^3 + y^3 + z^3 + 1 = 30$. The solutions satisfying both conditions are $(3,1,1)$, $(1,3,1)$, and $(1,1,3)$.
Correct Answer: 3