Questão 4 - 1ª Fase - IME 2027

$\mathrm{A} = \left|\begin{array}{ccc}{\mathrm{x}}^{\mathrm{m}}& {\mathrm{x}}^{\mathrm{m} + 2}& {\mathrm{x}}^{\mathrm{m} + 4}\\ {\mathrm{y}}^{\mathrm{m}}& {\mathrm{y}}^{\mathrm{m} + 2}& {\mathrm{y}}^{\mathrm{m} + 4}\\ {\mathrm{z}}^{\mathrm{m}}& {\mathrm{z}}^{\mathrm{m} + 2}& {\mathrm{z}}^{\mathrm{m} + 4}\end{array}\right| = {\mathrm{x}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}{\mathrm{z}}^{\mathrm{m}} \left|\begin{array}{ccc}1& {\mathrm{x}}^2& {\mathrm{x}}^4\\ 1& {\mathrm{y}}^2& {\mathrm{y}}^4\\ 1& {\mathrm{z}}^2& {\mathrm{z}}^4\end{array}\right| = {\mathrm{x}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}{\mathrm{z}}^{\mathrm{m}}({\mathrm{z}}^2 - {\mathrm{y}}^2)({\mathrm{z}}^2 - {\mathrm{x}}^2)({\mathrm{y}}^2 - {\mathrm{x}}^2)$

$\mathrm{B} = \left(\frac{1}{{\mathrm{y}}^2} - \frac{1}{{\mathrm{x}}^2}\right)\left(\frac{1}{{\mathrm{z}}^2} - \frac{1}{{\mathrm{y}}^2}\right)\left(\frac{1}{{\mathrm{x}}^2} - \frac{1}{{\mathrm{z}}^2}\right) = \frac{({\mathrm{x}}^2 - {\mathrm{y}}^2)({\mathrm{y}}^2 - {\mathrm{z}}^2)({\mathrm{z}}^2 - {\mathrm{x}}^2)}{{\mathrm{x}}^4{\mathrm{y}}^4{\mathrm{z}}^4}$

Como $\mathrm{A} = \mathrm{B}$, $\boxed{\mathrm{m} = -4}$.