Precalculus proving trigonometric identities

The LCM  is sinx*cos^3x....so we have

cos(x) cos^3 (x) / [ sin (x) * cos^3 (x)] +

[2 sinx * sin x * cos^2 (x)] / [sin (x)cos^3 (x)] +

[ sin (x) sin^3(x)] / [ sin(x) cos^3 (x) ]  =

[cos^4 (x) + 2 sin (x) cos^3(x) + sin^4 (x)] / [ sin (x)cos^3(x) ] =

[cos^4 (x) + 2 sin (x) cos(x) * cos^2(x) + sin^4 (x)] / [ sin (x)cos^3(x) ]  =

[ cos^4 (x) + sin(2x)*cos^2(x) + sin^4(x) ]  / [ sin(x)cos^3(x) ]

(cosx / sinx) + (2sinx / cosx) + (sin^3x / cos^3x)

How to simplify the expression with the LCM?

$\begin{array}{|rcll|} \hline && \frac{\cos(x)}{\sin(x)} + \frac{2\sin(x) }{ \cos(x) } + \frac{ \sin^3(x) } { \cos^3(x) } \\ &=& \frac{\cos(x)}{\sin(x)}\cdot \frac{\cos^3(x)}{\cos^3(x)} + \frac{2\sin(x) }{ \cos(x) } \cdot \frac{\sin(x)\cos^2(x)}{\sin(x)\cos^2(x)} + \frac{ \sin^3(x) } { \cos^3(x) }\cdot \frac{ \sin(x) } { \sin(x) } \\ &=& \frac{\cos(x)\cos^3(x)+ 2\sin(x)\sin(x)\cos^2(x) + \sin^3(x)\sin(x) } {\sin(x)\cos^3(x) } \\ &=& \frac{\cos^4(x)+ 2\sin^2(x)\cos^2(x) + \sin^4(x) } {\sin(x)\cos^3(x) } \\ &=& \frac{\sin^4(x)+2\sin^2(x)\cos^2(x)+\cos^4(x) } {\sin(x)\cos^3(x) } \\ &=& \frac{\Big( \sin^2(x)+\cos^2(x) \Big)^2 } {\sin(x)\cos^3(x) } \quad & | \quad \sin^2(x)+\cos^2(x) = 1 \\ &=& \frac{ 1^2 } {\sin(x)\cos^3(x) } \\ &=& \frac{ 1 } {\sin(x)\cos^3(x) } \\ \hline \end{array}$