$$(6^{-1}-8^{-1})^{-1}=$$
$24$
$$\left(\frac{1}{6}-8^{-1}\right)^{-1}$$
$$\left(\frac{1}{6}-\frac{1}{8}\right)^{-1}$$
$$\left(\frac{1}{24}\right)^{-1}$$
$$24$$
Show Solution
Hide Solution
$2^{3}\times 3$