Answer
See the explanation below.
Work Step by Step
Let us consider the left side of the given expression:
$\frac{{{\sin }^{3}}x-{{\cos }^{3}}x}{\sin x-\cos x}$
By using the common factor as $\sin x-\cos x$ in simplifying the identity, we get:
$\begin{align}
& \frac{{{\sin }^{3}}x-{{\cos }^{3}}x}{\sin x-\cos x}=\frac{\left( \sin x-\cos x \right)\left( {{\sin }^{2}}x+\sin x\cos x+{{\cos }^{2}}x \right)}{\sin x-\cos x} \\
& ={{\sin }^{2}}x+{{\cos }^{2}}x+\sin x\cos x \\
& =1+\sin x\cos x
\end{align}$
Hence, the left side of the expression is equal to the right side, which is $\frac{{{\sin }^{3}}x-{{\cos }^{3}}x}{\sin x-\cos x}=1+\sin x\cos x$.
