If $\sin \mathrm{\theta }-\cos \mathrm{\theta }=1$ then the value of ${\sin }^3\mathrm{\theta }-{\cos }^3\mathrm{\theta }$ is

$\begin{array}{l}\sin \theta -\cos \theta =1\\ \text{ or }{\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta =1\\ \begin{array}{cl}\sin \theta \cos \theta =0& \\ \text{ Now }{\sin }^3\theta -{\cos }^3\theta & =1\left({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta \right)\\ & =(\sin \theta -\cos \theta {)}^2+3\sin \theta \cos \theta \\ & =1+3\sin \theta \cos \theta =1\end{array}\end{array}$