$$\lim_{x\longrightarrow\ \frac{\pi}{4}}\ \frac{\tan^3x-\cot x}{\sin\left(x-\frac{\pi}{4}\right)}=$$

Substitute $$x = \frac{\pi}{4} + h$$, where $$h \to 0$$.

The denominator becomes:

$$\sin\left(\frac{\pi}{4} + h - \frac{\pi}{4}\right) = \sin h$$

For the numerator, let $$t = \tan h$$.

Using the expansion property for $$\tan\left(\frac{\pi}{4} + h\right) = \frac{1+t}{1-t}$$ and $$\cot\left(\frac{\pi}{4} + h\right) = \frac{1-t}{1+t}$$, we get:

$$\tan^3 x - \cot x = \left(\frac{1+t}{1-t}\right)^3 - \frac{1-t}{1+t}$$

$$= \frac{(1+t)^4 - (1-t)^4}{(1-t)^3(1+t)}$$

Using the algebraic expansion difference $$(1+t)^4 - (1-t)^4 = 8t + 8t^3 = 8t(1+t^2)$$:

$$\text{Numerator} = \frac{8t(1+t^2)}{(1-t)^3(1+t)}$$

Substitute this back into the limit expression:

$$\lim_{h \to 0} \frac{8 \tan h (1 + \tan^2 h)}{(1 - \tan h)^3(1 + \tan h)\sin h}$$

Separate the terms into standard limit forms:

$$= 8 \cdot \left(\lim_{h \to 0} \frac{\tan h}{\sin h}\right) \cdot \left(\lim_{h \to 0} \frac{1 + \tan^2 h}{(1 - \tan h)^3(1 + \tan h)}\right)$$

Since $$\lim_{h \to 0} \frac{\tan h}{\sin h} = 1$$ and as $$h \to 0$$, $$\tan h \to 0$$:

$$= 8 \times 1 \times \frac{1 + 0}{(1 - 0)^3(1 + 0)} = 8$$