Key Concept: Binomial Theorem and General Term

For a binomial expansion of the form $(a + b)^n$, the general term, denoted as $T_{r+1}$, which is the $(r+1)^{th}$ term from the beginning, is given by the formula: $T_{r+1} = {}^nC_r \cdot a^{n-r} \cdot b^r$ where ${}^nC_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

Step-by-step Solution

1. Identify the Parameters of the Binomial Expansion: The given binomial expression is ${\left( {{{{x^3}} \over 3} + {3 \over x}} \right)^8}$. Comparing this with $(a+b)^n$: $a = \frac{x^3}{3}$ $b = \frac{3}{x}$ $n = 8$

2. Determine the Middle Term(s) of the Expansion: The total number of terms in the expansion of $(a+b)^n$ is $n+1$. In this case, $n=8$, so there are $8+1 = 9$ terms. When $n$ is an even number, there is only one middle term, which is the $(\frac{n}{2} + 1)^{th}$ term. For $n=8$, the middle term is the $(\frac{8}{2} + 1)^{th} = (4+1)^{th} = 5^{th}$ term. Therefore, we need to find $T_5$. This means $r = 4$ for the general term formula $T_{r+1}$.

3. Write out the General Term for the Middle Term ($T_5$): Using the general term formula $T_{r+1} = {}^nC_r \cdot a^{n-r} \cdot b^r$ with $n=8$, $r=4$, $a = \frac{x^3}{3}$, and $b = \frac{3}{x}$: $T_5 = {}^8C_4 \cdot \left(\frac{x^3}{3}\right)^{8-4} \cdot \left(\frac{3}{x}\right)^4$ $T_5 = {}^8C_4 \cdot \left(\frac{x^3}{3}\right)^4 \cdot \left(\frac{3}{x}\right)^4$

4. Calculate the Binomial Coefficient: Now, calculate the value of ${}^8C_4$: ${}^8C_4 = \frac{8!}{4!(8-4)!} = \frac{8!}{4!4!}$ ${}^8C_4 = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4 \times 3 \times 2 \times 1 \times 4!}$ ${}^8C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}$ ${}^8C_4 = \frac{1680}{24} = 70$

5. Simplify the Middle Term: Substitute the value of ${}^8C_4$ back into the expression for $T_5$ and simplify the terms involving $x$: $T_5 = 70 \cdot \frac{(x^3)^4}{3^4} \cdot \frac{3^4}{x^4}$ $T_5 = 70 \cdot \frac{x^{12}}{81} \cdot \frac{81}{x^4}$ Notice that $3^4 = 81$, so the numerical denominators and numerators cancel out. $T_5 = 70 \cdot x^{12-4}$ $T_5 = 70x^8$

6. Equate the Middle Term to the Given Value and Solve for $x$: The problem states that the middle term equals 5670. $70x^8 = 5670$ To find $x$, first isolate $x^8$: $x^8 = \frac{5670}{70}$ $x^8 = \frac{567}{7}$ $x^8 = 81$ To solve for $x$, we need to find the 8th root of 81. We know that $3^4 = 81$. $x^8 = 3^4$ We can also express 81 as $(\sqrt{3})^8$, since $(\sqrt{3})^2 = 3$, and $(\sqrt{3})^8 = ((\sqrt{3})^2)^4 = 3^4 = 81$. So, $x^8 = (\sqrt{3})^8$ Taking the 8th root of both sides, remembering to include both positive and negative real roots for an even exponent: $x = \pm \sqrt{3}$

The sum of the real values of $x$ is $\sqrt{3} + (-\sqrt{3}) = 0$.

Tips and Common Mistakes:

• Counting Terms: Always remember that for an expansion of $(a+b)^n$, there are $n+1$ terms.
• Identifying Middle Term: If $n$ is even, there is one middle term $(\frac{n}{2} + 1)^{th}$. If $n$ is odd, there are two middle terms $(\frac{n+1}{2})^{th}$ and $(\frac{n+3}{2})^{th}$.
• Exponent Rules: Be careful when simplifying terms like $(x^3)^4 = x^{12}$ and when combining powers of $x$ in division ($\frac{x^{12}}{x^4} = x^{12-4} = x^8$).
• Solving for Even Powers: When solving equations like $x^8 = k$, always consider both positive and negative roots if $k > 0$.

Summary: We used the general term formula of the binomial expansion to identify and calculate the $5^{th}$ term, which is the middle term. After simplifying the expression for the middle term to $70x^8$, we equated it to the given value of 5670. Solving the resulting equation $x^8 = 81$ yielded the real values $x = \pm \sqrt{3}$. The sum of these real values is 0.