If $\dfrac {\sin^{3}\theta+\cos^{3}\theta}{\sin \theta+\cos \theta}+\dfrac {\sin^{3}\theta-\cos^{3}\theta}{\sin \theta-\cos \theta}=K$,  then K =

#### Solution

We have,

$\dfrac{{{\sin }^{3}}\theta +{{\cos }^{3}}\theta }{\sin \theta +\cos \theta }+\dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\sin \theta -\cos \theta }=K$

$\dfrac{\left( \sin \theta +\cos \theta \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta -\sin \theta \cos \theta \right)}{\left( \sin \theta +\cos \theta \right)}+\dfrac{\left( \sin \theta -\cos \theta \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +\sin \theta \cos \theta \right)}{\left( \sin \theta -\cos \theta \right)}=K$

$\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta -\sin \theta \cos \theta \right)+\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +\sin \theta \cos \theta \right)=K$

$1-\sin \theta \cos \theta +1+\sin \theta \cos \theta =K$

$K=2$

Hence, this is the answer.