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(A) The binomial expansion has $n=8$ terms,so the middle term is the $\left(\frac{8}{2} + 1ight)^{	ext{th}} = 5^{	ext{th}}$ term.The general term

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(A) The binomial expansion has $n=8$ terms,so the middle term is the $\left(\frac{8}{2} + 1ight)^{	ext{th}} = 5^{	ext{th}}$ term.The general term $t_{r+1}$ is given by $^{8}C_{r} \left(\frac{x^{3}}{3}ight)^{8-r} \left(\frac{3}{x}ight)^{r}$.For the $5^{	ext{th}}$ term,$r=4$:$t_{5} = ^{8}C_{4} \left(\frac{x^{3}}{3}ight)^{4} \left(\frac{3}{x}ight)^{4} = 5670$$^{8}C_{4} = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} = 70$.$70 	imes \frac{x^{12}}{3^{4}} 	imes \frac{3^{4}}{x^{4}} = 5670$$70 	imes x^{8} = 5670$$x^{8} = \frac{5670}{70} = 81$$x^{8} = 81 \implies x^{4} = 9$ or $x^{4} = -9$ (not possible for real $x$).$x^{4} = 9 \implies x^{2} = 3$ or $x^{2} = -3$ (not possible for real $x$).$x^{2} = 3 \implies x = \sqrt{3}$ or $x = -\sqrt{3}$.The sum of the real values of $x$ is $\sqrt{3} + (-\sqrt{3}) = 0$.

The sum of the real values of $x$ for which the middle term in the binomial expansion of ${\left( {\frac{{{x^3}}}{3} + \frac{3}{x}} \right)^8}$ equals $5670$ is

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