Let $$f : R - \{\frac{\alpha}{6}\} \to R$$ be defined by $$f(x) = \left(\frac{5x+3}{6x-\alpha}\right)$$. Then the value of $$\alpha$$ for which $$(f \circ f)(x) = x$$, for all $$x \in R - \{\frac{\alpha}{6}\}$$, is:

We have $$f(x) = \frac{5x + 3}{6x - \alpha}$$. We need $$(f \circ f)(x) = x$$ for all $$x \in R \setminus \{\frac{\alpha}{6}\}$$.

Computing $$f(f(x))$$ directly: let $$u = f(x) = \frac{5x+3}{6x-\alpha}$$, then $$f(u) = \frac{5u + 3}{6u - \alpha} = \frac{5\cdot\frac{5x+3}{6x-\alpha} + 3}{6\cdot\frac{5x+3}{6x-\alpha} - \alpha}.$$

Multiplying numerator and denominator by $$(6x - \alpha)$$: $$f(f(x)) = \frac{5(5x+3) + 3(6x-\alpha)}{6(5x+3) - \alpha(6x-\alpha)} = \frac{25x + 15 + 18x - 3\alpha}{30x + 18 - 6\alpha x + \alpha^2}.$$

For this to equal $$x$$ for all valid $$x$$, we need $$\frac{(25+18)x + (15-3\alpha)}{(30-6\alpha)x + (18+\alpha^2)} = x$$, i.e., $$(43)x + (15-3\alpha) = x\left[(30-6\alpha)x + (18+\alpha^2)\right].$$

For this to hold as an identity, the coefficient of $$x^2$$ on the right must be zero: $$30 - 6\alpha = 0$$, but this gives $$\alpha = 5$$. Let us verify: with $$\alpha = 5$$, the numerator becomes $$(25+18)x + (15-15) = 43x$$ and the denominator becomes $$(30-30)x + (18+25) = 43$$, so $$f(f(x)) = \frac{43x}{43} = x$$. ✓

Therefore $$\alpha = 5$$.