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

$a, b \neq 0, \quad a, b \in \mathbb{R}$ $\left(ax^{2} + \dfrac{70}{27bx}\right)^{4}$ $\left(ax - \dfrac{1}{bx^{2}}\right)^{7}$ $\text{Coeff of } x^{5} \text{ in } (1) = \text{Coeff of } x^{-5} \text{ in } (2)$ $\text{Find } 2b$ $\text{Plan: write } T_{r+1} \text{ for each, match the power of } x$ $\text{then equate the two coefficients}$ $(X+Y)^{n} = \sum_{r=0}^{n} \binom{n}{r} X^{n-r} Y^{r}$ $T_{r+1} = \binom{n}{r} X^{n-r} Y^{r}$ $T_{k+1} = \binom{4}{k} (ax^{2})^{4-k} \left(\dfrac{70}{27bx}\right)^{k}$ $= \binom{4}{k} a^{4-k}\!\left(\dfrac{70}{27b}\right)^{k} x^{\,8-3k}$ $8-3k = 5 \ \Rightarrow\ k = 1$ $\text{Coeff of } x^{5} = \binom{4}{1} a^{3}\,\dfrac{70}{27b} = \dfrac{280\,a^{3}}{27b}$ $T_{j+1} = \binom{7}{j} (ax)^{7-j} \left(-\dfrac{1}{bx^{2}}\right)^{j}$ $= \binom{7}{j} (-1)^{j} a^{7-j} b^{-j}\, x^{\,7-3j}$ $7-3j = -5 \ \Rightarrow\ j = 4$ $\text{Coeff of } x^{-5} = \binom{7}{4} a^{3} b^{-4} = \dfrac{35\,a^{3}}{b^{4}}$ $\dfrac{280\,a^{3}}{27b} = \dfrac{35\,a^{3}}{b^{4}}$ $\Rightarrow \dfrac{280}{27b} = \dfrac{35}{b^{4}} \quad (a \neq 0)$ $280\,b^{4} = 35 \cdot 27\,b$ $\Rightarrow 280\,b^{3} = 945 \ \Rightarrow\ b^{3} = \dfrac{945}{280} = \dfrac{27}{8}$ $b = \sqrt[3]{\dfrac{27}{8}} = \dfrac{3}{2}$ $2b = 3$ $\text{Coeff of } x^{5} = \dfrac{280\,a^{3}}{27\cdot\frac{3}{2}} = \dfrac{560\,a^{3}}{81}$ $\text{Coeff of } x^{-5} = \dfrac{35\,a^{3}}{\left(\frac{3}{2}\right)^{4}} = \dfrac{560\,a^{3}}{81}$