If the coefficient of x^{8} in left(a x^{2}+f | vedclass

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Question

(A) The general term in $\left(a x^{2}+\frac{1}{b x}\right)^{13}$ is $T_{r+1} = {}^{13}C_{r} (a x^{2})^{13-r} (b^{-1} x^{-1})^{r} = {}^{13}C_{r} a^{13

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$a b+1=0$

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(A) The general term in $\left(a x^{2}+\frac{1}{b x}\right)^{13}$ is $T_{r+1} = {}^{13}C_{r} (a x^{2})^{13-r} (b^{-1} x^{-1})^{r} = {}^{13}C_{r} a^{13-r} b^{-r} x^{26-3r}$.For the coefficient of $x^{8}$, set $26-3r = 8$, which gives $3r = 18$, so $r = 6$.The coefficient is ${}^{13}C_{6} a^{7} b^{-6}$.The general term in $\left(a x-\frac{1}{b x^{2}}\right)^{13}$ is $T'_{r+1} = {}^{13}C_{r} (a x)^{13-r} (-b^{-1} x^{-2})^{r} = {}^{13}C_{r} a^{13-r} (-1)^{r} b^{-r} x^{13-3r}$.For the coefficient of $x^{-8}$, set $13-3r = -8$, which gives $3r = 21$, so $r = 7$.The coefficient is ${}^{13}C_{7} a^{6} (-1)^{7} b^{-7} = -{}^{13}C_{7} a^{6} b^{-7}$.Equating the two coefficients:${}^{13}C_{6} a^{7} b^{-6} = -{}^{13}C_{7} a^{6} b^{-7}$.Since ${}^{13}C_{6} = {}^{13}C_{7}$, we have $a^{7} b^{-6} = -a^{6} b^{-7}$.Dividing by $a^{6} b^{-6}$, we get $a = -b^{-1}$, which implies $ab = -1$, or $ab+1=0$.

If the coefficient of $x^{8}$ in $\left(a x^{2}+\frac{1}{b x}\right)^{13}$ is equal to the coefficient of $x^{-8}$ in $\left(a x-\frac{1}{b x^{2}}\right)^{13},$ then $a$ and $b$ will satisfy the relation

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