If the coefficient of ${x^7}$ in ${\left( {ax - {1 \over {b{x^2}}}} \right)^{13}}$ and the coefficient of ${x^{ - 5}}$ in ${\left( {ax + {1 \over {b{x^2}}}} \right)^{13}}$ are equal, then ${a^4}{b^4}$ is equal to :

The given expression is $\left(a x-\frac{1}{b x^2}\right)^{13}$ <br/><br/>So, <br/><br/>$$ \begin{aligned} T_{r+1} & ={ }^{13} C_r(a x)^{13-r}\left(-\frac{1}{b x^2}\right)^r \\\\ & ={ }^{13} C_r(a)^{13-r} x^{13-r-2 r}(-1 / b)^r \\\\ & ={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r x^{13-3 r} \end{aligned} $$ <br/><br/>For coefficient of $x^7$ in $\left(a x-\frac{1}{b x^2}\right)^{13}$ <br/><br/>$$ \begin{aligned} & \quad 13-3 r=7 \\\\ & \Rightarrow 3 r=6 \Rightarrow r=2 \\\\ & \therefore \text { Coefficient of } x^7={ }^{13} C_2 \cdot(a)^{11} \cdot \frac{1}{b^2} \end{aligned} $$ <br/><br/>Again, the another expression is $\left(a x+\frac{1}{b x^2}\right)^{13}$ <br/><br/>So, $T_{n+1}={ }^{13} C_r(a x)^{13-r}\left(\frac{1}{b x^2}\right)^r={ }^{13} C_r(a)^{13-r}\left(\frac{1}{b}\right)^r x^{13-3 r}$ <br/><br/>For coefficient $x^{-5}$ in $\left(a x+\frac{1}{b x^2}\right)^{13}$ <br/><br/>$$ \begin{aligned} &13-3 r =-5 \\\\ &\Rightarrow r =6 \end{aligned} $$ <br/><br/>So, coefficient of $x^{-5}={ }^{13} C_6(a)^7 \frac{1}{b^6}$ <br/><br/>Now, according to the question, <br/><br/>${ }^{13} C_2(a)^{11} \frac{1}{b^2}={ }^{13} C_6(a)^7 \frac{1}{b^6}$ <br/><br/>$$ \begin{aligned} & \Rightarrow a^4 b^4=\frac{{ }^{13} C_6}{{ }^{13} C_2} \\\\ & \therefore a^4 b^4=22 \end{aligned} $$