To solve the problem, we will work step by step:

1. Interpret the sets:

• $A = \{1, 3, m\}$, where $m$ is a real number.
• $B = \{a^2 \mid a \in A\} = \{1^2, 3^2, m^2\} = \{1, 9, m^2\}$.
• $C = A \cup B$ means $C = \{1, 3, m, 9, m^2\}$ (since union combines elements without repetition).

2. Use the given condition:

The sum of the elements in $C$ is 20: $1 + 3 + m + 9 + m^2 = 20.$ Simplify: $m^2 + m + 13 = 20.$ $m^2 + m - 7 = 0.$

3. Solve for $m$:

Solve the quadratic equation $m^2 + m - 7 = 0$ using the quadratic formula: $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 1, b = 1, c = -7$: $m = \frac{-1 \pm \sqrt{1^2 - 4(1)(-7)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 28}}{2} = \frac{-1 \pm \sqrt{29}}{2}.$ Thus, the two possible values of $m$ are: $m = \frac{-1 + \sqrt{29}}{2} \quad \text{or} \quad m = \frac{-1 - \sqrt{29}}{2}.$

4. Compute the product of the elements in $C$:

The elements in $C$ are $\{1, 3, 9, m, m^2\}$. The product of these elements is: $P = 1 \cdot 3 \cdot 9 \cdot m \cdot m^2 = 27 \cdot m \cdot m^2.$ Simplify: $P = 27 \cdot m^3.$

5. Determine $m^3$:

From the quadratic equation $m^2 + m - 7 = 0$, we know: $m^2 = 7 - m.$ Substitute $m^2$ into $m^3 = m \cdot m^2$: $m^3 = m \cdot (7 - m) = 7m - m^2.$ Substitute $m^2 = 7 - m$ again: $m^3 = 7m - (7 - m) = 7m - 7 + m = 8m - 7.$

6. Compute the product $P$ for both cases:

• For $m = \frac{-1 + \sqrt{29}}{2}$: $m^3 = 8\left(\frac{-1 + \sqrt{29}}{2}\right) - 7 = 4(-1 + \sqrt{29}) - 7 = -4 + 4\sqrt{29} - 7 = -11 + 4\sqrt{29}.$ $P = 27 \cdot m^3 = 27(-11 + 4\sqrt{29}) = -297 + 108\sqrt{29}.$

• For $m = \frac{-1 - \sqrt{29}}{2}$: $m^3 = 8\left(\frac{-1 - \sqrt{29}}{2}\right) - 7 = 4(-1 - \sqrt{29}) - 7 = -4 - 4\sqrt{29} - 7 = -11 - 4\sqrt{29}.$ $P = 27 \cdot m^3 = 27(-11 - 4\sqrt{29}) = -297 - 108\sqrt{29}.$

Final Answer:

The product of the elements in $C$ is: $P = -297 + 108\sqrt{29} \quad \text{or} \quad P = -297 - 108\sqrt{29}.$

Would you like me to verify or simplify further?

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Relative Questions:

1. How does the quadratic formula work in solving equations like $m^2 + m - 7 = 0$?
2. Why does the union of sets $A$ and $B$ eliminate duplicate elements?
3. How can we determine whether both roots $m$ are valid solutions for this problem?
4. What role does the sum condition play in deriving the equation for $m$?
5. How does the choice of $m$ affect the product of the elements in $C$?

Tip: When working with sets and their elements, always verify whether duplicates are properly handled during operations like union