Question:

**First expansion** $\left(ax^2+\dfrac{70}{27bx}\right)^4$: general term $\binom{4}{r}(ax^2)^{4-r}\left(\dfrac{70}{27bx}\right)^r=\binom{4}{r}a^{4-r}\left(\dfrac{70}{27b}\right)^r x^{8-3r}$. For $x^5$: $8-3r=5\Rightarrow r=1$. Coefficient $=\binom{4}{1}a^3\cdot\dfrac{70}{27b}=\dfrac{280a^3}{27b}$. **Second expansion** $\left(ax-\dfrac{1}{bx^2}\right)^7$: general term $\binom{7}{r}(ax)^{7-r}\left(-\dfrac{1}{bx^2}\right)^r=\binom{7}{r}a^{7-r}\left(-\dfrac{1}{b}\right)^r x^{7-3r}$. For $x^{-5}$: $7-3r=-5\Rightarrow r=4$. Coefficient $=\binom{7}{4}a^3\cdot\dfrac{1}{b^4}=\dfrac{35a^3}{b^4}$. Setting equal: $\dfrac{280a^3}{27b}=\dfrac{35a^3}{b^4}\Rightarrow\dfrac{280}{27b}=\dfrac{35}{b^4}\Rightarrow b^3=\dfrac{35\cdot27}{280}=\dfrac{27}{8}\Rightarrow b=\dfrac{3}{2}$. $2b=3$.