Simplify: $\dfrac{{{\sin }^{3}}\theta +{{\cos }^{3}}\theta }{\sin \theta +\cos \theta }+\sin \theta \cos \theta $?

Hint: To simplify the given trigonometric expression, we are going to use the property of the algebraic expression which is as follows: ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$. This algebraic identity we are going to use in ${{\sin }^{3}}\theta +{{\cos }^{3}}\theta $. After applying this algebraic identity, $\left( \sin \theta +\cos \theta \right)$ will get cancelled out from the numerator and the denominator in the given expression. After that using simple algebra, we can solve the given expression.

Complete step by step answer:
The trigonometric expression given in the above problem is as follows:
$\dfrac{{{\sin }^{3}}\theta +{{\cos }^{3}}\theta }{\sin \theta +\cos \theta }+\sin \theta \cos \theta $
To solve the above expression, we are going to use the following algebraic identity which is equal to:
${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
We are going to apply the above identity in ${{\sin }^{3}}\theta +{{\cos }^{3}}\theta $. So, substituting $a=\sin \theta \And b=\cos \theta $ in the above equation we get,
${{\sin }^{3}}\theta +{{\cos }^{3}}\theta =\left( \sin \theta +\cos \theta \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta -\sin \theta \cos \theta \right)$
Now, applying the above equation in the given expression and we get,
$\dfrac{\left( \sin \theta +\cos \theta \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta -\sin \theta \cos \theta \right)}{\sin \theta +\cos \theta }+\sin \theta \cos \theta $
As you can see that in the above expression, $\left( \sin \theta +\cos \theta \right)$ is common in the numerator and the denominator so $\left( \sin \theta +\cos \theta \right)$ will get cancelled out and we are left with:
$\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta -\sin \theta \cos \theta \right)+\sin \theta \cos \theta $
Now, in the above expression, you can see that $\sin \theta \cos \theta $ will get cancelled out and the above expression will look like:
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta $
We also know the trigonometric identity which says that addition of square of sine and addition of square of cosine will give an answer as 1. The mathematical form of the above trigonometric identity will look like:
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$

Hence, the simplification of the expression given in the above problem is 1.

Note: To solve the above problem, you must know the algebraic identity of ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ and also the trigonometric identity which says that square of sine and square of cosine is 1. Failure to remember any of the identities inhibits you to solve the question further.