Let a and b be two nonzero real numbers. If the coefficient of $$x^5$$ in the expansion of $$\left(ax^2 + \frac{70}{27bx}\right)^4$$ is equal to the coefficient of $$x^{-5}$$ in the expansion of $$\left(ax - \frac{1}{bx^2}\right)^7$$, then the value of $$2b$$ is

Write the first expression in binomial form:
$$\left(ax^{2} + \frac{70}{27\,b\,x}\right)^{4}$$

General term of $$\left(u+v\right)^{4}$$ is $$\binom{4}{k}u^{k}v^{4-k}$$.
Take $$u = ax^{2}$$ and $$v = \dfrac{70}{27\,b\,x}$$.

The term containing $$k$$ copies of $$u$$ and $$4-k$$ copies of $$v$$ equals
$$\binom{4}{k}\,a^{k}\left(ax^{2}\right)^{k} \left(\frac{70}{27\,b\,x}\right)^{4-k} = \binom{4}{k}\,a^{k} \left(\frac{70}{27\,b}\right)^{4-k} x^{\,2k-(4-k)} = \binom{4}{k}\,a^{k} \left(\frac{70}{27\,b}\right)^{4-k} x^{\,3k-4}$$

We need the power of $$x$$ equal to $$5$$:
$$3k-4 = 5 \;\Longrightarrow\; 3k = 9 \;\Longrightarrow\; k = 3$$

Substituting $$k=3$$ gives the desired coefficient (call it $$C_1$$):
$$C_1 = \binom{4}{3}a^{3}\left(\frac{70}{27\,b}\right)^{1} = 4 \cdot a^{3}\cdot\frac{70}{27\,b} = \frac{280\,a^{3}}{27\,b}$$

Now treat the second expression:
$$\left(ax - \frac{1}{b\,x^{2}}\right)^{7}$$

Here $$u = ax$$, $$v = -\dfrac{1}{b\,x^{2}}$$ and the general term is
$$\binom{7}{r}u^{r}v^{7-r} = \binom{7}{r}a^{r}(ax)^{\,r} \left(-\frac{1}{b\,x^{2}}\right)^{7-r} = \binom{7}{r}a^{r}(-1)^{7-r}b^{-(7-r)} x^{\,r-2(7-r)} = \binom{7}{r}a^{r}(-1)^{7-r}b^{-(7-r)} x^{\,3r-14}$$

We need the power of $$x$$ equal to $$-5$$:
$$3r-14 = -5 \;\Longrightarrow\; 3r = 9 \;\Longrightarrow\; r = 3$$

With $$r=3$$, the power of $$-1$$ is $$(-1)^{7-3}=(-1)^{4}=+1$$.
The required coefficient (call it $$C_2$$) is
$$C_2 = \binom{7}{3}a^{3}b^{-(7-3)} = 35\,a^{3}\,b^{-4} = \frac{35\,a^{3}}{b^{4}}$$

According to the question, these two coefficients are equal:

$$\frac{280\,a^{3}}{27\,b} = \frac{35\,a^{3}}{b^{4}}$$

Cancel the common non-zero factor $$a^{3}$$ and cross-multiply:

$$280\,b^{3} = 35 \times 27$$

Divide by $$35$$:

$$b^{3} = \frac{27}{8}$$

Taking the real cube root (since $$b$$ is real and $$\dfrac{27}{8} \gt 0$$):

$$b = \frac{3}{2}$$

Finally, the required value is $$2b$$:

$$2b = 2 \times \frac{3}{2} = 3$$

Therefore, the answer is 3.