Key Concept: Binomial Expansion of Sum of Conjugate Terms

This problem utilizes a common simplification in binomial expansions involving terms of the form $(A+B)^n + (A-B)^n$. When two such expansions are added, all terms with odd powers of $B$ cancel out. This significantly reduces the number of terms we need to calculate.

The general formula for this type of expansion is: $(A+B)^n + (A-B)^n = 2 \left[ \binom{n}{0} A^n B^0 + \binom{n}{2} A^{n-2} B^2 + \binom{n}{4} A^{n-4} B^4 + \dots \right]$ This formula only includes terms where the power of $B$ is even.

Step-by-Step Solution

1. Identify the Components (A, B, n) Given the expression: ${\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}$ We can identify the components for our general formula:

   • $A = x$
   • $B = \sqrt{x^2-1}$
   • $n = 6$ Why this step? Breaking down the complex expression into its fundamental $A$, $B$, and $n$ components allows us to directly apply the powerful general formula for $(A+B)^n + (A-B)^n$.

2. Apply the Simplified Binomial Expansion Formula Substitute the identified values of $A$, $B$, and $n$ into the formula: $2 \left[ \binom{6}{0} x^6 (\sqrt{x^2-1})^0 + \binom{6}{2} x^4 (\sqrt{x^2-1})^2 + \binom{6}{4} x^2 (\sqrt{x^2-1})^4 + \binom{6}{6} x^0 (\sqrt{x^2-1})^6 \right]$ Why this step? By immediately using the simplified formula, we avoid expanding two full binomial expressions (each with 7 terms) and then combining them. This saves significant time and reduces the chance of errors.

3. Calculate Binomial Coefficients and Simplify Powers of $B$ Now, we calculate the binomial coefficients and simplify the terms involving $B$:

   • $\binom{6}{0} = 1$
   • $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$
   • $\binom{6}{4} = \binom{6}{6-4} = \binom{6}{2} = 15$ (using the property $\binom{n}{r} = \binom{n}{n-r}$)
   • $\binom{6}{6} = 1$

   And for the powers of $B = \sqrt{x^2-1}$:

   • $(\sqrt{x^2-1})^0 = 1$ (Any non-zero number raised to the power of 0 is 1)
   • $(\sqrt{x^2-1})^2 = x^2-1$
   • $(\sqrt{x^2-1})^4 = ((x^2-1)^{1/2})^4 = (x^2-1)^2$
   • $(\sqrt{x^2-1})^6 = ((x^2-1)^{1/2})^6 = (x^2-1)^3$

   Substitute these simplified values back into the expression: $2 \left[ 1 \cdot x^6 \cdot 1 + 15 \cdot x^4 (x^2-1) + 15 \cdot x^2 (x^2-1)^2 + 1 \cdot x^0 (x^2-1)^3 \right]$ $2 \left[ x^6 + 15x^4(x^2-1) + 15x^2(x^2-1)^2 + (x^2-1)^3 \right]$ Why this step? This step reduces the expression to a more manageable form, replacing the combinatorial notation and radical terms with their algebraic equivalents.

4. Expand and Simplify Algebraically Next, we expand the polynomial terms using algebraic identities:

   • $(x^2-1)^2 = (x^2)^2 - 2(x^2)(1) + 1^2 = x^4 - 2x^2 + 1$ (using $(a-b)^2 = a^2 - 2ab + b^2$)
   • $(x^2-1)^3 = (x^2)^3 - 3(x^2)^2(1) + 3(x^2)(1)^2 - 1^3 = x^6 - 3x^4 + 3x^2 - 1$ (using $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$)

   Substitute these expansions back into the expression: $2 \left[ x^6 + (15x^6 - 15x^4) + 15x^2(x^4 - 2x^2 + 1) + (x^6 - 3x^4 + 3x^2 - 1) \right]$ $2 \left[ x^6 + 15x^6 - 15x^4 + 15x^6 - 30x^4 + 15x^2 + x^6 - 3x^4 + 3x^2 - 1 \right]$ Why this step? Expanding these terms fully converts the expression into a standard polynomial form, making it easier to identify and collect coefficients of specific powers of $x$.

5. Collect Like Terms Now, group the terms by their powers of $x$:

   • Coefficient of $x^6$: $1x^6 + 15x^6 + 15x^6 + 1x^6 = (1+15+15+1)x^6 = 32x^6$
   • Coefficient of $x^4$: $-15x^4 - 30x^4 - 3x^4 = (-15-30-3)x^4 = -48x^4$
   • Coefficient of $x^2$: $15x^2 + 3x^2 = (15+3)x^2 = 18x^2$
   • Constant term: $-1$

   So, the expression inside the brackets simplifies to: $32x^6 - 48x^4 + 18x^2 - 1$ Why this step? Grouping terms by their powers allows us to clearly see the coefficient for each power of $x$, which is essential for determining $\alpha$ and $\beta$.

6. Complete the Expansion by Multiplying by 2 Finally, multiply the entire simplified polynomial by the factor of 2 that was extracted at the beginning: $2 \left[ 32x^6 - 48x^4 + 18x^2 - 1 \right] = 64x^6 - 96x^4 + 36x^2 - 2$ Why this step? This is the last step in getting the complete expanded form of the original expression, which is necessary to identify the exact coefficients.

Identify Coefficients $\alpha$ and $\beta$

From the final expanded expression $64x^6 - 96x^4 + 36x^2 - 2$:

• The coefficient of $x^4$ is $\alpha$. Therefore, $\alpha = -96$.
• The coefficient of $x^2$ is $\beta$. Therefore, $\beta = 36$. Why this step? This directly extracts the values of $\alpha$ and $\beta$ as defined in the problem statement.

Calculate the Required Expression

The problem asks us to find the relationship between $\alpha$ and $\beta$, specifically by checking the given options. Let's calculate both $\alpha + \beta$ and $\alpha - \beta$.

• $\alpha + \beta = -96 + 36 = -60$
• $\alpha - \beta = -96 - 36 = -132$

Match with Options

Comparing our results with the given options: (A) $\alpha + \beta = 60$ (Our result: -60) (B) $\alpha - \beta = 60$ (Our result: -132) (C) $\alpha + \beta = -30$ (Our result: -60) (D) $\alpha - \beta = -132$ (Our result: -132)

Our calculated value $\alpha - \beta = -132$ matches option (D).

Tips for Students & Common Mistakes

• Recognize the Sum/Difference Pattern: Always look for $(A+B)^n \pm (A-B)^n$. Knowing when to use the simplified formulas (only even powers of B for '+' and only odd powers of B for '-') is a significant time-saver.
• Careful with Negative Signs: A very common mistake is errors in handling negative signs, especially when expanding terms like $(x^2-1)^2$ or $(x^2-1)^3$ and when distributing constants. Double-check each sign change.
• Algebraic Identities: Memorizing and correctly applying identities like $(a-b)^2$ and $(a-b)^3$ is crucial for accuracy and speed.
• Systematic Approach: Break the problem into smaller, manageable steps: identify components, apply formula, calculate coefficients, expand algebraically, collect terms, then perform final calculations. This reduces the cognitive load and helps prevent errors.

Summary and Key Takeaway

This problem demonstrates the efficiency gained by recognizing specific patterns in binomial expansions. By leveraging the simplified formula for $(A+B)^n + (A-B)^n$, we methodically expanded the expression, carefully collected terms, and identified the coefficients of $x^4$ and $x^2$. The coefficients were found to be $\alpha = -96$ and $\beta = 36$, leading to $\alpha - \beta = -132$. The key takeaway is to look for binomial sum/difference patterns to simplify calculations and maintain precision through careful algebraic manipulation.