Let a and b be two nonzero real numbers. If the coefficient of x ^...

Given,

The coefficient of ${x}^{5}$ in ${\left(a{x}^{2} + \dfrac{70}{27bx}\right)}^{4}$ is equal to coefficient of ${x}^{- 5}$ in ${\left(ax - \dfrac{1}{b{x}^{2}}\right)}^{7}$,

Now finding the coefficient of ${x}^{5}$ in ${\left(a{x}^{2} + \dfrac{70}{27bx}\right)}^{4}$ we get,

${T}_{r + 1} = Cr 4 {\left(a{x}^{2}\right)}^{4 - r} \cdot {\left(\dfrac{70}{27bx}\right)}^{r}$

$⇒ {T}_{r + 1} = Cr 4 {\left(a\right)}^{4 - r} \cdot {\left(\dfrac{70}{27b}\right)}^{r} \cdot {x}^{8 - 2r - r}$

So, $8 - 2r - r = 5 ⇒ r = 1$

So, ${T}_{2} = C1 4 {a}^{3} \cdot \dfrac{70}{27b} \cdot {x}^{5}$

Now finding the coefficient of ${x}^{- 5}$ in ${\left(ax - \dfrac{1}{b{x}^{2}}\right)}^{7}$ we get,

${T}_{r + 1} = Cr 7 {\left(ax\right)}^{7 - r} \cdot {\left(\dfrac{1}{b{x}^{2}}\right)}^{r}$

$⇒ {T}_{r + 1} = Cr 7 {\left(a\right)}^{7 - r} \cdot {\left(\dfrac{1}{b}\right)}^{r} \cdot {x}^{7 - r - 2r}$

So, $7 - r - 2r = - 5 ⇒ r = 4$

So, ${T}_{5} = C4 7 {\left(a\right)}^{3} \cdot {\left(\dfrac{1}{b}\right)}^{4} \cdot {x}^{- 5}$

Now equating the coefficient of ${x}^{5} \text{ and } {x}^{- 5}$ we get,

$C4 7 {\left(a\right)}^{3} \cdot {\left(\dfrac{1}{b}\right)}^{4} = C1 4 {a}^{3} \cdot \dfrac{70}{27b}$

$⇒ 35{a}^{3} \cdot {\left(\dfrac{1}{b}\right)}^{4} = 4{a}^{3} \cdot \dfrac{70}{27b}$

$⇒ {\left(\dfrac{1}{b}\right)}^{3} = 4 \times \dfrac{2}{27}$

$⇒ b = \dfrac{3}{2} ⇒ 2b = 3$