If \(\ell < m < n\), then <br> <img src="https://latex.codecogs.com/svg.image?\begin{vmatrix} 1 & \ell & \ell^4 \\ 1 & m & m^4 \\ 1 & n & n^4 \end{vmatrix}" /> <br> will always be greater than -
Step-by-Step Solution
Key Concept: The determinant is a Vandermonde-like determinant. Specifically, it is (m-l)(n-l)(n-m)(l+m+n). Since l < m < n, all factors (m-l), (n-l), (n-m) are positive. The sign of the determinant depends on the sum (l+m+n). However, for positive l, m, n, the expression is positive.
The determinant is $D = (m-\ell)(n-\ell)(n-m)(\ell+m+n)$. Since $\ell < m < n$, we have $m-\ell > 0$, $n-\ell > 0$, and $n-m > 0$. Thus, the product is positive, meaning the determinant is greater than 0.
Correct Answer: 4