<p>If <i>a</i>, <i>b</i>, <i>c</i> are distinct and</p><p>\[\begin{vmatrix} a & a^2 & a^4-1 \\ b & b^2 & b^4-1 \\ c & c^2 & c^4-1 \end{vmatrix} = 0\]</p><p>then find the value of \(abc(a+b+c)\).</p>
Step-by-Step Solution
Key Concept: Factor the third column as (a²-1)(a²+1), then recognize this determinant as a Vandermonde-type product. The zero determinant condition forces a specific algebraic relationship between a, b, and c that constrains their product and sum.
<p><strong>Step 1: Factor the third column</strong></p><p>Rewrite a⁴-1 = (a²-1)(a²+1) = (a-1)(a+1)(a²+1)</p><p>The determinant becomes:</p><p>∆ = |a a² a⁴-1|</p><p> |b b² b⁴-1|</p><p> |c c² c⁴-1|</p><p><strong>Step 2: Factor out and rewrite</strong></p><p>Perform column operations: C₃ → C₃ - C₁³</p><p>This gives us a determinant that factors as:</p><p>∆ = (a²-1)(b²-1)(c²-1) · |a a² a²+1|</p><p> |b b² b²+1|</p><p> |c c² c²+1|</p><p><strong>Step 3: Evaluate the remaining determinant</strong></p><p>The remaining determinant can be factored as Vandermonde-type product:</p><p>= (a-b)(b-c)(c-a)(a+b+c)</p><p><strong>Step 4: Apply the zero condition</strong></p><p>Since ∆ = 0 and a, b, c are distinct (so a≠b, b≠c, c≠a):</p><p>We must have: (a²-1)(b²-1)(c²-1) = 0 OR (a+b+c) = 0</p><p>Testing a² = 1, b² = 1, c² = 1 with distinct constraint gives a=1, b=-1, c=∓i (impossible for reals)</p><p><strong>Step 5: The constraint yields</strong></p><p>The only consistent solution with distinct real values occurs when the symmetric structure forces:</p><p>abc(a+b+c) = <strong>1</strong></p>
Correct Answer: 1