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Step 1: Identify the general term in the binomial expansion
The binomial expansion of
$ \bigl(\frac{x^3}{3} + \frac{3}{x}\bigr)^8 $
has the general term (the $(r+1)^\text{th}$ term) given by:
$ T_{r+1} = \binom{8}{r} \Bigl(\frac{x^3}{3}\Bigr)^{8-r} \Bigl(\frac{3}{x}\Bigr)^r. $
Step 2: Determine the middle term
For an expansion with $n=8$, there are $(8+1)=9$ terms. The middle term is the 5th term, which corresponds to $r=4$ (since $r$ starts from 0).
Step 3: Write the middle term explicitly
Substitute $r=4$ into the general term formula:
$ T_5 = \binom{8}{4} \Bigl(\frac{x^3}{3}\Bigr)^{8-4} \Bigl(\frac{3}{x}\Bigr)^{4}. $
Simplify the exponents:
$ T_5 = \binom{8}{4} \Bigl(\frac{x^3}{3}\Bigr)^{4} \Bigl(\frac{3}{x}\Bigr)^{4}. $
Step 4: Simplify the expression for the middle term
First, compute $ \binom{8}{4} $:
$ \binom{8}{4} = 70. $
Next, combine like terms carefully:
$ \Bigl(\frac{x^3}{3}\Bigr)^{4} = \frac{x^{12}}{3^4} = \frac{x^{12}}{81}, \quad
\Bigl(\frac{3}{x}\Bigr)^{4} = \frac{3^4}{x^4} = \frac{81}{x^4}. $
Therefore,
$ T_5 = 70 \times \frac{x^{12}}{81} \times \frac{81}{x^4} = 70 \cdot x^{8}. $
Step 5: Set the middle term equal to 5670 and solve for x
Given $ T_5 = 5670 $, we have
$ 70 \, x^{8} = 5670. $
Divide both sides by 70:
$ x^{8} = \frac{5670}{70} = 81. $
Taking the 8th root of both sides:
$ x = \pm \sqrt{3}. $
(The real 8th root solutions are $ \pm \sqrt{3} $, since $81 = ( \sqrt{3} )^8.)$
Step 6: Find the sum of the real values of x
The two real values of $ x $ are $ +\sqrt{3} $ and $ -\sqrt{3} $. Their sum is:
$ \sqrt{3} + (-\sqrt{3}) = 0. $
Step 7: State the final answer
The sum of the real values of $ x $ for which the middle term equals 5670 is 0.
