Evaluate:
$${\left(\frac{\sin 23^{\circ }\cos 67^{\circ }+\cos 23^{\circ }\sin 67^{\circ }}{{\mathrm{cosec}}^{2}15^{\circ }-{\tan }^{2}75^{\circ }}\right)}^3$$

Consider the numerator:
$$\sin 23^{\circ }\cos 67^{\circ }+\cos 23^{\circ }\sin 67^{\circ }$$
Using the identity $\sin (A+B)=\sin A\cos B+\cos A\sin B$:
$$\sin (23^{\circ }+67^{\circ })=\sin 90^{\circ }=1$$
Consider the denominator:
$${\mathrm{cosec}}^{2}15^{\circ }-{\tan }^{2}75^{\circ }$$
Since $\tan 75^{\circ }=\tan (90^{\circ }-15^{\circ })=\cot 15^{\circ }$:
$${\mathrm{cosec}}^{2}15^{\circ }-{\cot }^{2}15^{\circ }$$
Using the identity ${\mathrm{cosec}}^2\theta -{\cot }^2\theta =1$:
$$1$$
Thus, the expression becomes:
$${\left(\frac{1}{1}\right)}^3=1^3=1$$