Let $$\alpha \gt 0, \beta \gt 0$$ be such that $$\alpha^3 + \beta^2 = 4$$. If the maximum value of the term independent of $$x$$ in the binomial expansion of $$\left(\alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}}\right)^{10}$$ is $$10k$$, then $$k$$ is equal to:

We have to study the binomial expansion of $$\left(\alpha\,x^{\tfrac19}+\beta\,x^{-\tfrac16}\right)^{10}$$ where $$\alpha\gt 0,\;\beta\gt 0$$ and they satisfy the relation $$\alpha^{3}+\beta^{2}=4\;.$$

The general term (the $$(r+1)$$-th term) of $$(u+v)^{10}$$ is given by the binomial theorem

$$T_{r}=\,{{}^{10}C_{r}}\;u^{\,10-r}\,v^{\,r}\;.$$

Here $$u=\alpha\,x^{\tfrac19},\;v=\beta\,x^{-\tfrac16}\;,$$ so

$$T_{r}={{}^{10}C_{r}}\left(\alpha\,x^{\tfrac19}\right)^{10-r}\!\left(\beta\,x^{-\tfrac16}\right)^{r} ={{}^{10}C_{r}}\;\alpha^{\,10-r}\beta^{\,r}\;x^{\;\tfrac{10-r}{9}-\tfrac{r}{6}}\;.$$

We want the term independent of $$x$$, that is, the exponent of $$x$$ must be zero:

$$\frac{10-r}{9}-\frac{r}{6}=0\;.$$

Multiplying by $$18$$ to clear denominators, we obtain

$$2(10-r)-3r=0\;\Longrightarrow\;20-2r-3r=0\;\Longrightarrow\;20-5r=0\;\Longrightarrow\;r=4\;.$$

Substituting $$r=4$$ in the general term gives the required constant term:

$$T_{4}={{}^{10}C_{4}}\;\alpha^{\,10-4}\beta^{\,4} ={{}^{10}C_{4}}\;\alpha^{6}\beta^{4}\;.$$

We know $${}^{10}C_{4}=210\;,$$ hence

$$T_{4}=210\,\alpha^{6}\beta^{4}\;.$$

To maximise this expression subject to the constraint $$\alpha^{3}+\beta^{2}=4$$, we set

$$u=\alpha^{3}\quad(u\gt 0),\quad v=\beta^{2}\quad(v\gt 0)\;.$$

The given condition becomes

$$u+v=4\;.$$

Also,

$$\alpha^{6}=u^{2},\qquad\beta^{4}=v^{2}\;,$$

so

$$T_{4}=210\,u^{2}v^{2}=210\,(uv)^{2}\;.$$

Under the fixed sum $$u+v=4$$, the product $$uv$$ attains its maximum when $$u=v$$ (AM-GM inequality or elementary calculus). Thus

$$u=v=2\quad\Longrightarrow\quad uv=2\times2=4\;.$$

Therefore the maximum value of the required term is

$$T_{4,\max}=210\,(4)^{2}=210\times16=3360\;.$$

The question states that this maximum equals $$10k$$, so

$$10k=3360\;\Longrightarrow\;k=\frac{3360}{10}=336\;.$$

Hence, the correct answer is Option A.