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

6

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8

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0

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4

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Solution

The correct option is C $0$
Given ${(\frac{x^{3}}{3} + \frac{3}{x})}^{8}$

General term $= T_{r + 1} =^{8} C_{r} {(\frac{x^{3}}{3})}^{8 - r} {(\frac{3}{x})}^{r}$

Since n is even then the middle term is when $r = \frac{n}{2}$

Here, $\frac{n}{2} = \frac{8}{2} = 4$

$T_{5} =^{8} C_{4} \frac{x^{12}}{81} \times \frac{81}{x^{4}} = 5670$

$\Rightarrow 70 x^{8} = 5670$

$\Rightarrow x = \pm \sqrt{3}$

Hence, The sum of the real values of x are $0$

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