Question

Evaluate the determinant: \[ \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix}. \]

Hint:

Use column operations and factorization of differences of cubes to simplify determinants involving powers.

Answer 1

We can use properties of determinants or expand along the first row: \[ D = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix}. \] Subtract the first column from the second and third columns: \[ \Rightarrow \begin{vmatrix} 1 & 0 & 0 \\ a & b - a & c - a \\ a^3 & b^3 - a^3 & c^3 - a^3 \end{vmatrix} = \begin{vmatrix} b - a & c - a \\ b^3 - a^3 & c^3 - a^3 \end{vmatrix}. \] Now, \[ D = 1 \times \begin{vmatrix} b - a & c - a \\ b^3 - a^3 & c^3 - a^3 \end{vmatrix}. \] Recall the factorization: \[ b^3 - a^3 = (b - a)(b^2 + ab + a^2), \] \[ c^3 - a^3 = (c - a)(c^2 + ac + a^2). \] Thus, \[ D = \begin{vmatrix} b - a & c - a \\ (b - a)(b^2 + ab + a^2) & (c - a)(c^2 + ac + a^2) \end{vmatrix}. \] Factor out \( (b - a) \) and \( (c - a) \) from rows: \[ D = (b - a)(c - a) \begin{vmatrix} 1 & 1 \\ b^2 + ab + a^2 & c^2 + ac + a^2 \end{vmatrix}. \] Calculate the \(2 \times 2\) determinant: \[ = (b - a)(c - a) \left[ (c^2 + ac + a^2) - (b^2 + ab + a^2) \right] \] \[ = (b - a)(c - a) \left( c^2 + ac - b^2 - ab \right). \] Rewrite inside: \[ c^2 - b^2 + a(c - b) = (c - b)(c + b) + a(c - b) = (c - b)(c + b + a). \] So, \[ D = (b - a)(c - a)(c - b)(c + b + a). \] Note the order: To match the common Vandermonde pattern, reorder factors and account for sign changes: \[ D = (b - a)(c - a)(c - b)(a + b + c). \] \ Final answer: \[ \boxed{ \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} = (b - a)(c - a)(c - b)(a + b + c). } \]