Prove that for \(a,b,x,y > 0\),

\[ \frac{2}{3}\tan^{-1}\left(\frac{3ab^2 – a^3}{b^3 – 3a^2b}\right) + \frac{2}{3}\tan^{-1}\left(\frac{3xy^2 – x^3}{y^3 – 3x^2y}\right) = \tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2 – \beta^2}\right) \]
where \( \alpha = -ax + by,\; \beta = bx + ay \)

Solution:

Use the identity:

\[ \tan(3\theta) = \frac{3\tan\theta – \tan^3\theta}{1 – 3\tan^2\theta} \]

This implies:

\[ \tan^{-1}\left(\frac{3t – t^3}{1 – 3t^2}\right) = 3\tan^{-1}(t) \]

Rewrite given expressions:

\[ \tan^{-1}\left(\frac{3ab^2 – a^3}{b^3 – 3a^2b}\right) = \tan^{-1}\left(\frac{3(a/b) – (a/b)^3}{1 – 3(a/b)^2}\right) = 3\tan^{-1}\left(\frac{a}{b}\right) \]

\[ \tan^{-1}\left(\frac{3xy^2 – x^3}{y^3 – 3x^2y}\right) = 3\tan^{-1}\left(\frac{x}{y}\right) \]

Thus LHS becomes:

\[ \frac{2}{3}(3\tan^{-1}(a/b)) + \frac{2}{3}(3\tan^{-1}(x/y)) \]

\[ = 2\tan^{-1}\left(\frac{a}{b}\right) + 2\tan^{-1}\left(\frac{x}{y}\right) \]

\[ = 2\left[\tan^{-1}\left(\frac{a}{b}\right) + \tan^{-1}\left(\frac{x}{y}\right)\right] \]

Using identity:

\[ \tan^{-1}p + \tan^{-1}q = \tan^{-1}\left(\frac{p+q}{1 – pq}\right) \]

\[ = \tan^{-1}\left(\frac{ay + bx}{by – ax}\right) \]

Thus,

\[ \text{LHS} = 2\tan^{-1}\left(\frac{\beta}{-\alpha}\right) \]

Now use:

\[ 2\tan^{-1}(t) = \tan^{-1}\left(\frac{2t}{1 – t^2}\right) \]

\[ = \tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2 – \beta^2}\right) \]

Hence proved.

Final Answer:

\[ \frac{2}{3}\tan^{-1}\left(\frac{3ab^2 – a^3}{b^3 – 3a^2b}\right) + \frac{2}{3}\tan^{-1}\left(\frac{3xy^2 – x^3}{y^3 – 3x^2y}\right) = \tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2 – \beta^2}\right) \]