Question

Simplify: \(\tan^{-1}\!\left(\dfrac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}}\right)\).

Hint:

Divide numerator and denominator by a^3 and match against the tan(3 theta) expansion formula.

Answer 1

Step 1: Substitution x = a tan(phi):
Let \(x=a\tan\phi\). Then numerator \(3a^2x-x^3=a^3(3\tan\phi-\tan^3\phi)\) and denominator \(a^3-3ax^2=a^3(1-3\tan^2\phi)\).

Step 2: Cancelling a^3 and using the identity:
The ratio becomes \(\dfrac{3\tan\phi-\tan^3\phi}{1-3\tan^2\phi}=\tan(3\phi)\) directly by the standard triple-angle tangent formula.

Step 3: Undoing the inverse tangent:
\(\tan^{-1}(\tan3\phi)=3\phi\), and since \(\phi=\tan^{-1}(x/a)\), the result follows.

Final Answer:
\[ \boxed{3\tan^{-1}(x/a)} \]