If $$\frac{1^3 + 2^3 + 3^3 + \ldots \text{upto n terms}}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \ldots \text{upto n terms}} = \frac{9}{5}$$ then the value of $$n$$ is

We start with the equation $$\frac{1^3 + 2^3 + 3^3 + \cdots + n^3}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \cdots \text{ (n terms)}} = \frac{9}{5}$$.

Using the well-known formula for the sum of cubes, one finds that

$$1^3 + 2^3 + \cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$$.

On the other hand, the general term of the denominator is $$k(2k+1)$$ for $$k = 1,2,\ldots,n$$, so

$$\sum_{k=1}^{n} k(2k+1) = 2\sum_{k=1}^{n}k^2 + \sum_{k=1}^{n}k$$

which becomes

$$2\cdot\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$$

and hence

$$\frac{n(n+1)}{6}\bigl[2(2n+1)+3\bigr] = \frac{n(n+1)(4n+5)}{6}$$.

Substituting these results into the original equation gives

$$\frac{\left[\frac{n(n+1)}{2}\right]^2}{\frac{n(n+1)(4n+5)}{6}} = \frac{9}{5}$$

which simplifies to

$$\frac{n^2(n+1)^2}{4}\times\frac{6}{n(n+1)(4n+5)} = \frac{9}{5} \quad\Longrightarrow\quad \frac{3n(n+1)}{2(4n+5)} = \frac{9}{5}$$.

Cross-multiplying leads to

$$15n(n+1) = 18(4n+5)$$

or equivalently

$$15n^2 + 15n = 72n + 90 \quad\Longrightarrow\quad 15n^2 - 57n - 90 = 0$$.

Dividing by 3 simplifies this to

$$5n^2 - 19n - 30 = 0$$

and application of the quadratic formula gives

$$n = \frac{19 \pm \sqrt{361+600}}{10} = \frac{19 \pm \sqrt{961}}{10} = \frac{19 \pm 31}{10}$$.

Since $$n$$ must be a positive integer, one obtains

$$n = \frac{19 + 31}{10} = \frac{50}{10} = 5$$.

Therefore, the correct answer is 5.