Problem Statement: We need to find the sum of the coefficients of all even degree terms in $x$ in the expansion of ${{\left( {x + \sqrt {{x^3} - 1} } \right)}^6} + {{\left( {x - \sqrt {{x^3} - 1} } \right)}^6}$, where $x > 1$.

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1. Key Binomial Theorem Property

The problem involves the sum of two binomial expansions that are nearly identical, differing only by a sign. For any two terms $P$ and $Q$, and a positive integer $n$, the sum of their binomial expansions can be simplified using the following property: ${(P+Q)^n + (P-Q)^n = 2 \left[ \binom{n}{0} P^n Q^0 + \binom{n}{2} P^{n-2} Q^2 + \binom{n}{4} P^{n-4} Q^4 + \dots \right]}$ Explanation: When $(P+Q)^n$ and $(P-Q)^n$ are expanded, the terms involving odd powers of $Q$ will have opposite signs and thus cancel out when added. The terms involving even powers of $Q$ will have the same sign and will be doubled. This significantly reduces the number of terms we need to consider.

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2. Applying the Property to the Given Expression

In our problem, let's identify $P$, $Q$, and $n$:

• $P = x$
• $Q = \sqrt{x^3 - 1}$
• $n = 6$

Substituting these into the property, we get: {{\left( {x + \sqrt {{x^3} - 1} } \right)}^6} + {{\left( {x - \sqrt {{x^3} - 1} } \right)}^6} = 2 \left[ \binom{6}{0} x^6 (\sqrt{x^3-1})^0 + \binom{6}{2} x^4 (\sqrt{x^3-1})^2 + \binom{6}{4} x^2 (\sqrt{x^3-1})^4 + \binom{6}{6} x^0 (\sqrt{x^3-1})^6 \right]}

Explanation: We've replaced $P$ with $x$ and $Q$ with $\sqrt{x^3-1}$. Notice that $(\sqrt{x^3-1})^k$ simplifies nicely for even values of $k$. Specifically, $(\sqrt{x^3-1})^2 = x^3-1$, $(\sqrt{x^3-1})^4 = (x^3-1)^2$, and $(\sqrt{x^3-1})^6 = (x^3-1)^3$. Also, we need to recall the binomial coefficients:

• $\binom{6}{0} = 1$
• $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$
• $\binom{6}{4} = \frac{6 \times 5}{2 \times 1} = 15$
• $\binom{6}{6} = 1$

Substituting these values, the expression becomes: $2 \left[ 1 \cdot x^6 (1) + 15 \cdot x^4 (x^3-1) + 15 \cdot x^2 (x^3-1)^2 + 1 \cdot (x^3-1)^3 \right]$ $= 2 \left[ x^6 + 15x^4(x^3-1) + 15x^2(x^3-1)^2 + (x^3-1)^3 \right]$

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3. Expanding Terms and Identifying Coefficients of Even Degree $x$

Now we will expand each term inside the bracket and collect the coefficients of $x$ that have an even degree (i.e., $x^0, x^2, x^4, \dots$). The condition $x>1$ ensures that $\sqrt{x^3-1}$ is a real number, making the expansion straightforward.

Term 1: $x^6$ This is already in its simplest form.

• The degree of $x$ is $6$ (even).
• Coefficient: $1$

Term 2: $15x^4(x^3-1)$ Expand this term by distributing $15x^4$: $15x^4(x^3-1) = 15x^7 - 15x^4$

• For $15x^7$, the degree of $x$ is $7$ (odd). We ignore this for now.
• For $-15x^4$, the degree of $x$ is $4$ (even).
• Coefficient: $-15$

Term 3: $15x^2(x^3-1)^2$ First, expand $(x^3-1)^2$ using the formula $(a-b)^2 = a^2 - 2ab + b^2$: $(x^3-1)^2 = (x^3)^2 - 2(x^3)(1) + 1^2 = x^6 - 2x^3 + 1$ Now, multiply by $15x^2$: $15x^2(x^6 - 2x^3 + 1) = 15x^8 - 30x^5 + 15x^2$

• For $15x^8$, the degree of $x$ is $8$ (even).
• Coefficient: $15$
• For $-30x^5$, the degree of $x$ is $5$ (odd). We ignore this.
• For $15x^2$, the degree of $x$ is $2$ (even).
• Coefficient: $15$

Term 4: $(x^3-1)^3$ Expand this term using the formula $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$: $(x^3-1)^3 = (x^3)^3 - 3(x^3)^2(1) + 3(x^3)(1)^2 - 1^3$ $= x^9 - 3x^6 + 3x^3 - 1$

• For $x^9$, the degree of $x$ is $9$ (odd). We ignore this.
• For $-3x^6$, the degree of $x$ is $6$ (even).
• Coefficient: $-3$
• For $3x^3$, the degree of $x$ is $3$ (odd). We ignore this.
• For $-1$, this is a constant term, which can be considered $-1 \cdot x^0$. The degree of $x$ is $0$ (even).
• Coefficient: $-1$

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4. Summing the Coefficients of Even Degree Terms

Now, we collect all the identified coefficients of even degree terms from the expansions above:

1. From $x^6$: $1$
2. From $-15x^4$: $-15$
3. From $15x^8$: $15$
4. From $15x^2$: $15$
5. From $-3x^6$: $-3$
6. From $-1$ (constant term $x^0$): $-1$

Sum of these coefficients (inside the main bracket): $1 + (-15) + 15 + 15 + (-3) + (-1)$ $= 1 - 15 + 15 + 15 - 3 - 1$ $= (1 - 15 + 15) + (15 - 3 - 1)$ $= 1 + 11$ $= 12$

Finally, we must remember that the entire expression is multiplied by $2$. So, the total sum of coefficients of all even degree terms in $x$ is: $2 \times 12 = 24$

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5. Important Considerations and Tips

• Understanding "Even Degree Terms": It's crucial to understand that "even degree terms in $x$" refers to terms like $Cx^0, Cx^2, Cx^4, \dots$. The constant term ($Cx^0$) is included as $x^0$ is an even degree.
• Alternative Method (and why it might be tricky here): For a general polynomial $P(x)$, the sum of coefficients of even degree terms is $\frac{P(1) + P(-1)}{2}$. While this method is powerful, it requires $P(x)$ to be a polynomial in $x$. If we tried to apply this directly here, evaluating $P(-1)$ would involve $\sqrt{(-1)^3 - 1} = \sqrt{-2} = i\sqrt{2}$, leading to complex numbers. Since the question asks for the sum of coefficients in the expansion (implying the resulting polynomial form), the direct expansion method used here avoids complex numbers and ensures we are capturing the coefficients of the polynomial in $x$.
• Careful Expansion: Binomial expansions, especially with nested terms like $(x^3-1)^n$, require meticulous attention to detail to avoid algebraic errors.

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6. Conclusion

By utilizing the binomial theorem property for the sum of two expansions, we first simplified the given expression. Then, we meticulously expanded each resulting term, carefully identifying and collecting the coefficients of all powers of $x$ that have an even degree. Summing these coefficients yielded a total of $12$ inside the bracket, and multiplying by the factor of $2$ gave the final answer.

The sum of the coefficients of all even degree terms in $x$ in the given expansion is $24$.

The final answer is $\boxed{\text{24}}$.