Value of Coefficient Relationship in Binomial Expansions

To find the value of $2b$, we determine the general terms for both binomial expansions.

For the expansion of $\left( ax^2 + \frac{70}{27bx} \right)^4$, the general term is given by: $T_{r+1} = \binom{4}{r} (ax^2)^{4-r} \left( \frac{70}{27bx} \right)^r = \binom{4}{r} a^{4-r} \left(\frac{70}{27b}\right)^r x^{8-3r}$ Setting $8-3r = 5$ gives $r = 1$, yielding the coefficient of $x^5$ as $\frac{280 a^3}{27 b}$.

For the expansion of $\left( ax - \frac{1}{bx^2} \right)^7$, the general term is given by: $T_{k+1} = \binom{7}{k} (ax)^{7-k} \left( -\frac{1}{bx^2} \right)^k = \binom{7}{k} (-1)^k a^{7-k} b^{-k} x^{7-3k}$ Setting $7-3k = -5$ gives $k = 4$, yielding the coefficient of $x^{-5}$ as $\frac{35 a^3}{b^4}$.

Equating both coefficients: $\frac{280 a^3}{27 b} = \frac{35 a^3}{b^4}$

Since $a \neq 0$ and $b \neq 0$, this simplifies to: $8 b^3 = 27 \implies 2b = 3$