Key Concept: Properties of Binomial Expansion

This problem utilizes a common property derived from the Binomial Theorem, specifically for expressions of the form $(A+B)^n + (A-B)^n$.

The Binomial Theorem states: $(A+B)^n = \binom{n}{0}A^n + \binom{n}{1}A^{n-1}B + \binom{n}{2}A^{n-2}B^2 + \dots + \binom{n}{n-1}AB^{n-1} + \binom{n}{n}B^n$ $(A-B)^n = \binom{n}{0}A^n - \binom{n}{1}A^{n-1}B + \binom{n}{2}A^{n-2}B^2 - \dots + (-1)^{n-1}\binom{n}{n-1}AB^{n-1} + (-1)^n\binom{n}{n}B^n$

When we add these two expansions, all terms containing odd powers of $B$ cancel out due to their opposing signs. This simplifies the expression to: $(A+B)^n + (A-B)^n = 2 \left[ \binom{n}{0}A^nB^0 + \binom{n}{2}A^{n-2}B^2 + \binom{n}{4}A^{n-4}B^4 + \dots \right]$ This formula significantly reduces the number of terms we need to compute.

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Step-by-step Working with Explanations

1. Identify A, B, and n The given expression is: ${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}$ Comparing this to the general form $(A+B)^n + (A-B)^n$, we can identify:

• $A = x$
• $B = \sqrt{x^3 - 1}$
• $n = 5$

2. Apply the Simplified Binomial Sum Formula Substitute $A=x$, $B=\sqrt{x^3-1}$, and $n=5$ into the formula derived above. Since $n=5$, we will include terms where the power of $B$ is an even number up to 4 (because the maximum power of $B$ is $n=5$, and the next even power after 4 would be 6, which exceeds $n$). ${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5} = 2 \left[ \binom{5}{0}x^5(\sqrt{x^3-1})^0 + \binom{5}{2}x^{5-2}(\sqrt{x^3-1})^2 + \binom{5}{4}x^{5-4}(\sqrt{x^3-1})^4 \right]$ Explanation: We use this simplified formula to efficiently combine the two expansions, cancelling out the terms with odd powers of $\sqrt{x^3-1}$.

3. Evaluate Binomial Coefficients Let's compute the binomial coefficients for $n=5$:

• $\binom{5}{0} = 1$
• $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$
• $\binom{5}{4} = \frac{5 \times 4 \times 3 \times 2}{4 \times 3 \times 2 \times 1} = 5$ (Alternatively, $\binom{n}{k} = \binom{n}{n-k}$, so $\binom{5}{4} = \binom{5}{1} = 5$) Explanation: Calculating these values is a fundamental step in binomial expansion.

4. Evaluate Powers of B Next, we evaluate the powers of $B = \sqrt{x^3-1}$:

• $(\sqrt{x^3-1})^0 = 1$ Explanation: Any non-zero base raised to the power of zero is $1$.
• $(\sqrt{x^3-1})^2 = x^3-1$ Explanation: Squaring a square root cancels the root.
• $(\sqrt{x^3-1})^4 = ((\sqrt{x^3-1})^2)^2 = (x^3-1)^2$ Explanation: We can rewrite the fourth power as a square of a square, which simplifies the calculation. Now, expand $(x^3-1)^2$ using the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$: $(x^3-1)^2 = (x^3)^2 - 2(x^3)(1) + (1)^2 = x^6 - 2x^3 + 1$

5. Substitute and Simplify the Entire Expression Now, substitute the calculated binomial coefficients and powers of $B$ back into the formula from Step 2: $= 2 \left[ 1 \cdot x^5 \cdot 1 + 10 \cdot x^3 (x^3-1) + 5 \cdot x (x^6 - 2x^3 + 1) \right]$ $= 2 \left[ x^5 + (10x^3 \cdot x^3 - 10x^3 \cdot 1) + (5x \cdot x^6 - 5x \cdot 2x^3 + 5x \cdot 1) \right]$ $= 2 \left[ x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x \right]$ Explanation: We perform the multiplication and distribution of terms to expand the expression into a polynomial form.

6. Rearrange and Distribute the Factor of 2 To clearly identify the degree of each term, it's helpful to arrange the polynomial in descending powers of $x$: $= 2 \left[ 5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x \right]$ Now, distribute the outside factor of 2 to every term inside the bracket: $= 10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x$ Explanation: Rearranging terms helps in systematic identification of odd-degree terms. Distributing the constant multiplier ensures that the coefficients are fully determined before the final step.

7. Identify Odd Degree Terms and Sum their Coefficients The problem asks for the sum of the coefficients of all odd degree terms. We look for terms where the exponent of $x$ is an odd number. From the simplified polynomial $10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x$:

• Term $10x^7$: Degree is 7 (odd), Coefficient is $10$.
• Term $2x^5$: Degree is 5 (odd), Coefficient is $2$.
• Term $-20x^3$: Degree is 3 (odd), Coefficient is $-20$.
• Term $10x$: Degree is 1 (odd), Coefficient is $10$. Explanation: We meticulously go through each term of the polynomial and identify those where the power of $x$ is an odd number.

Sum of coefficients of odd degree terms = $10 + 2 + (-20) + 10$ Sum $= 10 + 2 - 20 + 10$ Sum $= 22 - 20$ Sum $= 2$

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Tips for Success & Common Mistakes to Avoid:

• Recognize the Pattern Immediately: The structure $(A+B)^n + (A-B)^n$ is a strong indicator to use the simplified binomial sum formula. This saves significant time and reduces errors compared to expanding both terms fully and then adding.
• Careful with Signs in Expansion: If you were to expand without the simplified formula, remember that the terms in $(A-B)^n$ alternate in sign.
• Algebraic Precision: Errors often occur during the algebraic simplification, especially when multiplying terms like $x^3(x^3-1)$ or expanding $(x^3-1)^2$. Double-check each multiplication and subtraction.
• Definition of "Odd Degree": Remember that a term like $5x^1$ has an odd degree (degree 1). A constant term (e.g., $C$ or $Cx^0$) has an even degree (degree 0).

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Summary/Key Takeaway: This problem is an excellent illustration of how understanding and applying specific properties derived from the Binomial Theorem can simplify complex calculations. By recognizing the form $(A+B)^n + (A-B)^n$, we efficiently expanded the expression, identifying and summing the coefficients of all odd-degree terms. The key was a systematic approach to polynomial manipulation and careful attention to algebraic detail.