Key Concept: Binomial Theorem for Sum of Expansions

The problem involves the sum of two binomial expansions of the form $(A+B)^n + (A-B)^n$. A crucial simplification exists for such expressions, which significantly reduces the computational effort.

For an integer $n$, the expansion of $(A+B)^n$ is given by: $(A+B)^n = \sum_{k=0}^{n} {}^nC_k A^{n-k} B^k$ And the expansion of $(A-B)^n$ is: $(A-B)^n = \sum_{k=0}^{n} {}^nC_k A^{n-k} (-B)^k$ When we add these two expansions, the terms where $k$ is odd will cancel out because $(-B)^k = -B^k$ for odd $k$. The terms where $k$ is even will be doubled because $(-B)^k = B^k$ for even $k$. Thus, for $n=5$: $(A+B)^5 + (A-B)^5 = 2 \left[ {}^5C_0 A^5 B^0 + {}^5C_2 A^3 B^2 + {}^5C_4 A^1 B^4 \right]$ This formula is the foundation of our solution.

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Step 1: Expanding the Given Expression

We are given the expression: $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5$ Here, we can identify $A=x$ and $B=\sqrt{x^3-1}$. Applying the simplified binomial expansion formula for $(A+B)^5 + (A-B)^5$: $= 2\left({}^5C_0 x^5 (\sqrt{x^3-1})^0 + {}^5C_2 x^3 (\sqrt{x^3-1})^2 + {}^5C_4 x^1 (\sqrt{x^3-1})^4\right)$ Explanation: We use the simplified formula to avoid expanding all 12 terms (6 for each binomial) and then canceling. This saves time and reduces chances of error. We only include terms where the power of $\sqrt{x^3-1}$ is even.

Now, we calculate the binomial coefficients and simplify the powers of $B$:

• ${}^5C_0 = 1$
• ${}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$
• ${}^5C_4 = {}^5C_{5-4} = {}^5C_1 = 5$
• $(\sqrt{x^3-1})^0 = 1$
• $(\sqrt{x^3-1})^2 = x^3-1$
• $(\sqrt{x^3-1})^4 = (x^3-1)^2$

Substituting these values back into the expression: $= 2\left(1 \cdot x^5 \cdot 1 + 10 \cdot x^3 (x^3-1) + 5 \cdot x (x^3-1)^2\right)$ $= 2\left(x^5 + 10x^3(x^3-1) + 5x(x^6 - 2x^3 + 1)\right)$ Explanation: We performed the direct substitutions and expanded $(x^3-1)^2$ as a quadratic.

Further simplifying by distributing terms: $= 2\left(x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x\right)$ Rearranging the terms in descending powers of $x$: $= 2\left(5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x\right)$ Explanation: We multiplied through by $x^3$ and $x$ respectively and then arranged the terms for easier identification of coefficients.

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Step 2: Identifying the Coefficients $\alpha, \beta, \gamma, \delta$

The problem defines $\alpha, \beta, \gamma, \delta$ as the coefficients of $x^7, x^5, x^3$ and $x$ respectively. From our expanded expression: $2\left(5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x\right)$

• Coefficient of $x^7$: $\alpha = 2 \times 5 = 10$
• Coefficient of $x^5$: $\beta = 2 \times 1 = 2$
• Coefficient of $x^3$: $\gamma = 2 \times (-10) = -20$
• Coefficient of $x$: $\delta = 2 \times 5 = 10$

Explanation: We carefully extract the coefficient for each specified power of $x$ from the fully simplified expansion. Remember to multiply by the common factor of $2$ that was outside the bracket.

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Step 3: Setting Up and Solving the System of Linear Equations

We are given two linear equations:

1. $\alpha u+\beta v=18$
2. $\gamma u+\delta v=20$

Substitute the identified coefficients into these equations:

1. $10u + 2v = 18$
2. $-20u + 10v = 20$

To make these equations simpler, we can divide each by their greatest common divisor:

1. Divide by 2: $5u + v = 9$ (Equation A)
2. Divide by 10: $-2u + v = 2$ (Equation B)

Explanation: Simplifying the equations by dividing by common factors makes them easier to work with and reduces the chance of arithmetic errors during solving.

Now, we solve this system of two linear equations for $u$ and $v$. A straightforward method is elimination: Subtract Equation B from Equation A: $(5u + v) - (-2u + v) = 9 - 2$ $5u + v + 2u - v = 7$ $7u = 7$ $u = 1$ Explanation: Subtracting the equations eliminates the variable $v$, allowing us to directly solve for $u$.

Now substitute the value of $u=1$ into Equation A: $5(1) + v = 9$ $5 + v = 9$ $v = 9 - 5$ $v = 4$ Explanation: Substituting the value of $u$ back into one of the simplified equations allows us to find the value of $v$.

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Step 4: Calculating $u+v$

Finally, we need to find the value of $u+v$: $u+v = 1+4 = 5$

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Tips and Common Mistakes:

• Binomial Expansion Shortcut: Always look for patterns like $(A+B)^n \pm (A-B)^n$. Using the simplified expansion $2\left[ {}^nC_0 A^n + {}^nC_2 A^{n-2}B^2 + \dots \right]$ for the sum and $2\left[ {}^nC_1 A^{n-1}B^1 + {}^nC_3 A^{n-3}B^3 + \dots \right]$ for the difference can save significant time and prevent errors.
• Careful with Signs: Pay close attention to signs, especially when terms like $B^k$ involve negative expressions or when distributing the factor of $2$ in the final step. A common mistake is to forget to multiply the identified coefficients by the outer factor of $2$.
• Algebraic Simplification: Ensure that all powers are correctly combined (e.g., $x^3 \cdot x^3 = x^6$) and that the terms are fully expanded before identifying coefficients.
• System of Equations: Double-check your arithmetic when solving simultaneous equations. Small errors here can propagate and lead to an incorrect final answer.

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Summary and Key Takeaway:

This problem effectively tests the application of the Binomial Theorem, particularly the sum of two binomials with conjugate terms. It then combines this with the ability to extract specific coefficients from a polynomial expansion and solve a system of linear equations. The key takeaway is to leverage the symmetry in binomial expansions (i.e., using $(A+B)^n + (A-B)^n = 2 \sum_{k \text{ even}} {}^nC_k A^{n-k} B^k$) to simplify calculations and then meticulously follow through with algebraic manipulation and solving linear equations.