Key Concept: Rationalization and Binomial Theorem This problem requires two main mathematical concepts:

1. Rationalization: Simplifying expressions by eliminating square roots from the denominator. This is achieved by multiplying the numerator and denominator by the conjugate of the denominator, using the identity $(a-b)(a+b) = a^2 - b^2$.
2. Binomial Theorem: Expanding expressions of the form $(A+B)^N$. A useful property for this problem is the sum of two binomial expansions: $(A+B)^N + (A-B)^N = 2 \left[ \binom{N}{0} A^N + \binom{N}{2} A^{N-2} B^2 + \binom{N}{4} A^{N-4} B^4 + \dots \right]$ This formula helps in quickly identifying that only terms with even powers of $B$ (and corresponding even powers of $N-k$ for $A$) will remain and be doubled, while terms with odd powers of $B$ cancel out.

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Step 1: Simplify the Terms by Rationalizing the Denominator The given expression is: ${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8} + {\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$

Let's focus on the fraction inside the first bracket: ${{{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}}}$. To rationalize its denominator, we multiply the numerator and denominator by its conjugate, which is ${\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }$. Let $A = \sqrt {5{x^3} + 1}$ and $B = \sqrt {5{x^3} - 1}$. The denominator is $A-B$. Multiplying by $\frac{A+B}{A+B}$: ${{2 \over {A - B}} \times {{A + B} \over {A + B}}} = {{2(A + B)} \over {A^2 - B^2}}$ Now, calculate $A^2 - B^2$: $A^2 - B^2 = (\sqrt {5{x^3} + 1})^2 - (\sqrt {5{x^3} - 1})^2 = (5x^3 + 1) - (5x^3 - 1) = 5x^3 + 1 - 5x^3 + 1 = 2$ Substituting this back into the simplified fraction: ${{2(A + B)} \over 2} = A + B = \sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1}$ So, the first term of the original expression becomes: ${\left( {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} } \right)^8}$

Next, let's simplify the fraction inside the second bracket: ${{{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}}}$. Similarly, we multiply the numerator and denominator by its conjugate, which is ${\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }$. Using $A = \sqrt {5{x^3} + 1}$ and $B = \sqrt {5{x^3} - 1}$, the denominator is $A+B$. Multiplying by $\frac{A-B}{A-B}$: ${{2 \over {A + B}} \times {{A - B} \over {A - B}}} = {{2(A - B)} \over {A^2 - B^2}}$ As calculated before, $A^2 - B^2 = 2$. Substituting this back into the simplified fraction: ${{2(A - B)} \over 2} = A - B = \sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1}$ So, the second term of the original expression becomes: ${\left( {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} } \right)^8}$

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Step 2: Combine the Simplified Terms Now, the entire given expression transforms into a sum of two binomials raised to the power of 8: ${\left( {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} } \right)^8} + {\left( {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} } \right)^8}$ To make the application of the binomial theorem clearer, let's set: $X = \sqrt{5x^3 + 1}$ $Y = \sqrt{5x^3 - 1}$ The expression is now in the form: $(X+Y)^8 + (X-Y)^8$.

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Step 3: Apply the Binomial Theorem Using the identity mentioned in the key concept for $N=8$: $(X+Y)^8 + (X-Y)^8 = 2 \left[ \binom{8}{0} X^8 + \binom{8}{2} X^6 Y^2 + \binom{8}{4} X^4 Y^4 + \binom{8}{6} X^2 Y^6 + \binom{8}{8} Y^8 \right]$ Now, we substitute back the values of $X$ and $Y$. Note that $X^2 = (\sqrt{5x^3 + 1})^2 = 5x^3 + 1$ and $Y^2 = (\sqrt{5x^3 - 1})^2 = 5x^3 - 1$. Substitute these squares into the expansion: $2 \left[ \binom{8}{0} (X^2)^4 + \binom{8}{2} (X^2)^3 (Y^2)^1 + \binom{8}{4} (X^2)^2 (Y^2)^2 + \binom{8}{6} (X^2)^1 (Y^2)^3 + \binom{8}{8} (Y^2)^4 \right]$ $2 \left[ \binom{8}{0} (5x^3 + 1)^4 + \binom{8}{2} (5x^3 + 1)^3 (5x^3 - 1)^1 + \binom{8}{4} (5x^3 + 1)^2 (5x^3 - 1)^2 + \binom{8}{6} (5x^3 + 1)^1 (5x^3 - 1)^3 + \binom{8}{8} (5x^3 - 1)^4 \right]$

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Step 4: Determine the Degree of the Polynomial (n) The degree of the polynomial is the highest power of $x$ present in any of its terms. Let's examine the power of $x$ in each term within the large bracket: Each term is of the form $C \cdot (5x^3 + 1)^j (5x^3 - 1)^k$, where $j+k=4$. When we expand $(5x^3 + 1)^j$, the highest power of $x$ comes from $(5x^3)^j = 5^j x^{3j}$. When we expand $(5x^3 - 1)^k$, the highest power of $x$ comes from $(5x^3)^k = 5^k x^{3k}$. When these are multiplied, the highest power of $x$ in that term will be $x^{3j} \cdot x^{3k} = x^{3(j+k)}$. Since $j+k=4$ for all terms in the sum (e.g., for the first term $j=4, k=0$; for the second term $j=3, k=1$, etc.), the highest power of $x$ in any term will be $x^{3 \times 4} = x^{12}$. Therefore, the degree of the polynomial, $n = 12$.

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Step 5: Determine the Coefficient of $x^n$ (m) We need to find the coefficient of $x^{12}$. From Step 4, we observed that the $x^{12}$ term in each part of the expansion comes from multiplying the $(5x^3)^j$ part of $(5x^3 + 1)^j$ with the $(5x^3)^k$ part of $(5x^3 - 1)^k$. So, the coefficient of $x^{12}$ from each term of the form $(5x^3 + 1)^j (5x^3 - 1)^k$ (where $j+k=4$) will be $5^j \cdot 5^k = 5^{j+k} = 5^4$.

Now, let's sum these coefficients, multiplied by their respective binomial coefficients from the outer expansion: $m = 2 \left[ \binom{8}{0} (5^4) + \binom{8}{2} (5^4) + \binom{8}{4} (5^4) + \binom{8}{6} (5^4) + \binom{8}{8} (5^4) \right]$ We can factor out $5^4$: $m = 2 \cdot 5^4 \left[ \binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8} \right]$ Recall the binomial identity for the sum of even-indexed binomial coefficients: $\sum_{k \text{ even}} \binom{N}{k} = 2^{N-1}$. For $N=8$, the sum inside the bracket is $2^{8-1} = 2^7 = 128$. So, the coefficient $m$ is: $m = 2 \cdot 5^4 \cdot 2^7$ $m = 2^{1} \cdot 5^4 \cdot 2^7 = 2^{1+7} \cdot 5^4 = 2^8 \cdot 5^4$ We can express this further: $m = (2^2)^4 \cdot 5^4 = 4^4 \cdot 5^4 = (4 \times 5)^4 = 20^4$ Calculating the numerical value: $20^4 = (20 \times 20) \times (20 \times 20) = 400 \times 400 = 160000$.

Thus, the degree of the polynomial is $n=12$, and the coefficient of $x^n$ is $m=20^4$. The ordered pair $(n, m)$ is $(12, 20^4)$.

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Tips and Common Mistakes to Avoid:

• Don't skip rationalization: Always simplify expressions with square roots in the denominator first. It usually makes the rest of the problem much more manageable.
• Recognize Binomial Identities: The sum of even-indexed binomial coefficients ($2^{N-1}$) is a powerful shortcut.
• Careful with Powers of Powers: Remember that $(x^a)^b = x^{ab}$. This is crucial when determining the overall degree of the polynomial.
• Identify the Correct Coefficient: Ensure you are extracting the coefficient of the exact power of $x$ requested, not just the base of $x^3$.

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Summary: We meticulously simplified the given expression by rationalizing the denominators, transforming it into a sum of two binomials of the form $(X+Y)^8 + (X-Y)^8$. By applying the Binomial Theorem property for such sums, we expanded the expression. We then systematically identified that the highest power of $x$ would be $x^{12}$, thus determining the degree $n=12$. Finally, we calculated the coefficient of $x^{12}$ by summing the relevant binomial coefficients and powers of 5, which yielded $m = 20^4$. The final answer is the ordered pair $(12, 20^4)$.