Using the trigonometric identity for tan(A + B), prove that tan(3x)=(3tan(x)-tan^3(x))/(1-3tan^2(x))

tan(3x)=tan(2x+x), by using the identity for tan(A+B)=(tan(A)+tan(B))/(1-tan(A)tan(B)),tan(3x)=tan(2x+x)=(tan(2x)+tan(x))/(1-tan(2x)tan(x)), using it again for tan(2x),tan(3x)=tan(2x+x)=([(tan(x)+tan(x))/(1-tan(x)tan(x))]+tan(x))/(1-[(tan(x)+tan(x))/(1-tan(x)tan(x))]tan(x))which simplifies to ([2tan(x)/(1-tan²(x))]+tan(x))/(1-[(2tan(x))/(1-tan²(x))]tan(x))which will further simplify to [(3tan(x)+tan³(x))/(1-tan²(x))]/[(1-3tan²(x))/(1-tan²(x))]which yields [3tan(x)+tan³(x)]/[1-3tan²(x)] which is what we got asked
link to resolution on paper: https://imgur.com/a/YUuaop9