Let A be the 3x3 matrix whose (i,j) entry is (1 + a^(i-1)*b^(j-1) + a^(2(i-1))*b^(2(j-1))) evaluated at positions using values a, b, c as row/column parameters: A = [[1+a²+a⁴, 1+ab+a²*b², 1+ac+a²*c²], [1+ab+a²*b², 1+b²+b⁴, 1+bc+b²*c²], [1+ac+a²*c², 1+bc+b²*c², 1+c²+c⁴]]. If det(A) = det(4I

Question

Let A be the 3x3 matrix whose (i,j) entry is (1 + a^(i-1)*b^(j-1) + a^(2(i-1))*b^(2(j-1))) evaluated at positions using values a, b, c as row/column parameters: A = [[1+a²+a⁴, 1+ab+a²*b², 1+ac+a²*c²], [1+ab+a²*b², 1+b²+b⁴, 1+bc+b²*c²], [1+ac+a²*c², 1+bc+b²*c², 1+c²+c⁴]]. If det(A) = det(4I) where I is the 3x3 identity matrix, find |(a-b)³ + (b-c)³ + (c-a)³|.

Accepted Answer

24. Each entry A_(ij) = 1 + x_i*y_j + x_i²*y_j² where x = (1,a,a²) corresponds to... actually the entry (i,j) seems to use row parameter p_i in {1,a,a²} and column parameter q_j in {1,b,c} etc. The matrix can be written as A = P * Q^T where P has rows [1, p, p²] and Q has rows [1, q, q²] (Vandermonde-type). det(A) = det(P)*det(Q). The Vandermonde determinant structure gives det(A) = [(a-1)(b-1)(c-1)(b-a)(c-a)(c-b)]² or similar. det(4I) = 64. Using identity (a-b)³+(b-c)³+(c-a)³ = 3(a-b)(b-c)(c-a) when (a-b)+(b-c)+(c-a)=0 (which is always true). If det(A) = 64 leads to (a-b)²*(b-c)²*(c-a)² = k, then |(a-b)³+(b-c)³+(c-a)³| = |3(a-b)(b-c)(c-a)| = 3*|(a-b)(b-c)(c-a)| = 3*8 = 24.

Solution

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