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

Binomial Theorem PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q31 with Solution

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Correct answer: A

For the expansion of $\left(\frac{x^3}{3}+\frac{3}{x}\right)^8,$ there are $8+1=9$ terms.

Since the number of terms is odd, the middle term is the $\left(\frac{8}{2}+1\right)=5$ th term.

For $(a+b)^8$ , the general term is $T_{r+1}=\binom{8}{r}a^{8-r}b^r.$

Here, $a=\frac{x^3}{3}, \qquad b=\frac{3}{x}.$

So the 5th term corresponds to $r=4$ : $T_5=\binom{8}{4}\left(\frac{x^3}{3}\right)^4\left(\frac{3}{x}\right)^4.$

First, $\binom{8}{4}=70.$

Now, $\left(\frac{x^3}{3}\right)^4=\frac{x^{12}}{3^4}= rac{x^{12}}{81},$ $\left(\frac{3}{x}\right)^4=\frac{3^4}{x^4}=\frac{81}{x^4}.$

Multiplying,

=\frac{x^{12}}{81}\cdot \frac{81}{x^4} =x^8.$$ Hence, $$T_5=70x^8.$$ --- 4. **Use the given condition** The middle term equals $5670$, so $$70x^8=5670.$$ Thus, $$x^8=\frac{5670}{70}=81.$$ So, $$x=\pm 81^{1/8}.$$ Since $$81=3^4,$$ we get $$81^{1/8}=3^{4/8}=3^{1/2}=\sqrt{3}.$$ Therefore the real values are $$x=\sqrt{3},\quad x=-\sqrt{3}.$$ --- 5. **Find their sum** $$\sqrt{3}+(-\sqrt{3})=0.$$ --- 6. **Evaluate options** - **A: 0** ✅ - **B: 8** ❌ - **C: 6** ❌ - **D: 4** ❌ So the correct option is **A**.