Question

sin^3x+cos^3x over sinx+cosx = 1-sinx cosx

Answer 1

\[\frac{\sin^3x+\cos^3x}{sinx+cosx}\]
\[\sin^3x+\cos^3x=(sinx+cosx)(\sin^2x-sinx.cosx+\cos^2x)\]
so\[=\frac{\cancel{(sinx+cosx)}(\sin^2x-sinxcosx+\cos^2x)}{\cancel{sinx+cosx}}\]
\[ \sin^2x-sinxcosx+\cos^2x\] \[=1-sinxcosx\]