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.