Giải thích các bước giải:

a.Ta có:

$\dfrac{\sin^3\alpha+\cos^3\alpha}{\sin\alpha+\cos\alpha}$

$=\dfrac{(\sin\alpha+\cos\alpha)(\sin^2\alpha-\sin\alpha\cos\alpha+\cos^2\alpha)}{\sin\alpha+\cos\alpha}$

$=\sin^2\alpha-\sin\alpha\cos\alpha+\cos^2\alpha$

$=\sin^2\alpha+\cos^2\alpha-\sin\alpha\cos\alpha$

$=1-\sin\alpha\cos\alpha$

b.Ta có:

$\dfrac{1+\tan\alpha}{1-\tan\alpha}+\dfrac{1+\cot\alpha}{1-\cot\alpha}$

$=\dfrac{\tan\alpha\cot\alpha+\tan\alpha}{\tan\alpha\cot\alpha-\tan\alpha}+\dfrac{1+\cot\alpha}{1-\cot\alpha}$

$=\dfrac{\cot\alpha+1}{\cot\alpha-1}-\dfrac{\cot\alpha+1}{\cot\alpha-1}$

$=0$