The value of $\left(\frac{x}{y}\right)^{2a-3b}×\left(\frac{x}{y}\right)^{3b-4c}×\left(\frac{x}{y}\right)^{4c-2a}$

$\frac{x}{y}$ 1 0 $\left(\frac{x}{y}\right)^{2a+3b+4c}$

1

$\left(\frac{x}{y}\right)^{2a-3b+3b-4c+4c-2a}=\left(\frac{x}{y}\right)^0=1$