Prove that: `tan^3x/(1-3tan^2x)=1/8(tan3x-3tanx)`

 a. Prove that: `tan^3x/(1-3tan^2x)=1/8(tan3x-3tanx)`

As we knew: `tan3x=(3tanx-tan^3x)/(1-3tan^2x)`

                                  `=(3tanx-9tan^3x+8tan^3x)/(1-3tan^2x)`

                                  `=(8tan^3x+3tanx(1-3tan^2x))/(1-3tan^2x)`

                                  `=(8tan^3x)/(1-3tan^2x)+3tanx`

Then,             `tan3x-3tanx=(8tan^3x)/(1-3tan^2x)`

Hence,         `(tan^3x)/(1-3tan^2x)=1/8(tan3x-3tanx)`

b. Find the sum of `S_n=\sum_{k=1}^{n}(3^ktan^3(a/3^k))/(1-3tan^2(a/3^k))`

From our premise above: `(tan^3x)/(1-3tan^2x)=1/8(tan3x-3tanx)`

Then,    `(3^ktan^3(a/3^k))/(1-3tan^2(a/3^k))=3^k/8(tan(a/3^(k-1))-3tan(a/3^k)`

`k=1\rightarrow(3tan^3(a/3))/(1-3tan^2(a/3))=3^1/8(tana-3tan(a/3))`

`k=2\rightarrow(3^2tan^3(a/3^2))/(1-3tan^2(a/3^2))=3^2/8(tan(a/3)-3tan(a/3^2))`

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`k=n\rightarrow(3^ntan^3(a/3^n))/(1-3tan^2(a/3^n))=3^n/8(tan(a/3^(n-1))-3tan(a/3^n))`

Sum side to side:

`S_n=3/8tana-3^(n+1)/8tan(a/3^n)=3/8(tana-3^ntan(a/3^n))`