Step 1: Application of the Multinomial Expansion Formula

To expand the expression $(1 + \frac{x}{2} - \frac{2}{x})^{4}$, we treat the trinomial as a binomial by grouping terms: $[1 + (\frac{x}{2} - \frac{2}{x})]^{4}$.

Per the Binomial Theorem, for any positive integer $n$:

$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Setting $a = 1$, $b = (\frac{x}{2} - \frac{2}{x})$, and $n = 4$:

$(1 + \frac{x}{2} - \frac{2}{x})^4 = \binom{4}{0}(1)^4 + \binom{4}{1}(1)^3(\frac{x}{2} - \frac{2}{x}) + \binom{4}{2}(1)^2(\frac{x}{2} - \frac{2}{x})^2 + \binom{4}{3}(1)^1(\frac{x}{2} - \frac{2}{x})^3 + \binom{4}{4}(\frac{x}{2} - \frac{2}{x})^4$

Step 2: Evaluating the Coefficients and Sub-expansions

We calculate the binomial coefficients: $\binom{4}{0}=1, \binom{4}{1}=4, \binom{4}{2}=6, \binom{4}{3}=4, \binom{4}{4}=1$.

• Term 1: $1 \cdot 1 = 1$
• Term 2: $4(\frac{x}{2} - \frac{2}{x}) = 2x - \frac{8}{x}$
• Term 3: $6(\frac{x^2}{4} - 2 \cdot \frac{x}{2} \cdot \frac{2}{x} + \frac{4}{x^2}) = 6(\frac{x^2}{4} - 2 + \frac{4}{x^2}) = \frac{3x^2}{2} - 12 + \frac{24}{x^2}$
• Term 4: $4(\frac{x}{2} - \frac{2}{x})^3 = 4[(\frac{x}{2})^3 - 3(\frac{x}{2})^2(\frac{2}{x}) + 3(\frac{x}{2})(\frac{2}{x})^2 - (\frac{2}{x})^3]$
  $= 4[\frac{x^3}{8} - \frac{3x}{2} + \frac{6}{x} - \frac{8}{x^3}] = \frac{x^3}{2} - 6x + \frac{24}{x} - \frac{32}{x^3}$
• Term 5: $(\frac{x}{2} - \frac{2}{x})^4 = (\frac{x}{2})^4 - 4(\frac{x}{2})^3(\frac{2}{x}) + 6(\frac{x}{2})^2(\frac{2}{x})^2 - 4(\frac{x}{2})(\frac{2}{x})^3 + (\frac{2}{x})^4$
  $= \frac{x^4}{16} - 4(\frac{x^3}{8})(\frac{2}{x}) + 6(\frac{x^2}{4})(\frac{4}{x^2}) - 4(\frac{x}{2})(\frac{8}{x^3}) + \frac{16}{x^4} = \frac{x^4}{16} - x^2 + 6 - \frac{16}{x^2} + \frac{16}{x^4}$

Step 3: Summation and Algebraic Simplification

Summing all components calculated in Step 2:

Power of x	Summation
$x^4$	$\frac{x^4}{16}$
$x^3$	$\frac{x^3}{2}$
$x^2$	$\frac{3x^2}{2} - x^2 = \frac{x^2}{2}$
$x^1$	$2x - 6x = -4x$
Constant	$1 - 12 + 6 = -5$
$x^{-1}$	$-\frac{8}{x} + \frac{24}{x} = \frac{16}{x}$
$x^{-2}$	$\frac{24}{x^2} - \frac{16}{x^2} = \frac{8}{x^2}$
$x^{-3}$	$-\frac{32}{x^3}$
$x^{-4}$	$\frac{16}{x^4}$

Final Solution: \frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x - 5 + \frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4}

Given Expression & Initial Setup

To find the expansion of $(3x^2 - 2ax + 3a^2)^3$, we utilize the Binomial Theorem, which states $(A + B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k} B^k$. Since the expression is a trinomial, we group it into a binomial form: $[(3x^2 + 3a^2) - 2ax]^3$.

Step 1: Application of the Binomial Identity

Let $A = (3x^2 + 3a^2)$ and $B = -2ax$. Per the Binomial Theorem for $n=3$:

$(A + B)^3 = 1 \cdot A^3 + 3 \cdot A^2 \cdot B + 3 \cdot A \cdot B^2 + 1 \cdot B^3$

Step 2: Substitution and Expansion of Terms

Substituting $A = 3(x^2 + a^2)$ and $B = -2ax$:

1. Term 1 ($A^3$): $[3(x^2 + a^2)]^3 = 27(x^2 + a^2)^3$

Using $(x^2+a^2)^3 = (x^2)^3 + 3(x^2)^2(a^2) + 3(x^2)(a^2)^2 + (a^2)^3 = x^6 + 3x^4a^2 + 3x^2a^4 + a^6$.

Result: $27x^6 + 81x^4a^2 + 81x^2a^4 + 27a^6$.

2. Term 2 ($3A^2B$): $3[3(x^2+a^2)]^2 \cdot (-2ax) = 3[9(x^4 + 2x^2a^2 + a^4)] \cdot (-2ax)$

$= 27(x^4 + 2x^2a^2 + a^4) \cdot (-2ax) = -54ax^5 - 108a^3x^3 - 54a^5x$.

3. Term 3 ($3AB^2$): $3[3(x^2+a^2)] \cdot (-2ax)^2 = 9(x^2+a^2) \cdot (4a^2x^2)$

$= 36a^2x^2(x^2+a^2) = 36x^4a^2 + 36x^2a^4$.

4. Term 4 ($B^3$): $(-2ax)^3 = -8a^3x^3$.

Step 3: Aggregation and Algebraic Simplification

Summing the calculated terms:

Component	Algebraic Expression
$A^3$	$27x^6 + 81x^4a^2 + 81x^2a^4 + 27a^6$
$3A^2B$	$-54x^5a - 108x^3a^3 - 54xa^5$
$3AB^2$	$36x^4a^2 + 36x^2a^4$
$B^3$	$-8x^3a^3$

Combining like terms by degree:

• $x^6$ term: $27x^6$
• $x^5$ term: $-54ax^5$
• $x^4$ term: $(81 + 36)x^4a^2 = 117x^4a^2$
• $x^3$ term: $(-108 - 8)x^3a^3 = -116x^3a^3$
• $x^2$ term: $(81 + 36)x^2a^4 = 117x^2a^4$
• $x^1$ term: $-54xa^5$
• Constant ($x^0$) term: $27a^6$

Final Solution: $27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6$

Step 1: Establishing the Binomial Expansion Structure

To evaluate the expression $(a^{2} + \sqrt{a^{2}-1})^{4} + (a^{2} - \sqrt{a^{2}-1})^{4}$, we denote $x = a^{2}$ and $y = \sqrt{a^{2}-1}$. The expression simplifies to the form $(x + y)^{4} + (x - y)^{4}$.

By the Binomial Theorem, the expansion for $(x + y)^n$ is $\sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$. Thus:

$(x + y)^4 = \binom{4}{0}x^4 + \binom{4}{1}x^3y + \binom{4}{2}x^2y^2 + \binom{4}{3}xy^3 + \binom{4}{4}y^4$ [Per Binomial Theorem]

$(x - y)^4 = \binom{4}{0}x^4 - \binom{4}{1}x^3y + \binom{4}{2}x^2y^2 - \binom{4}{3}xy^3 + \binom{4}{4}y^4$ [By substitution of $-y$ for $y$ in the expansion]

Step 2: Combining the Expansions

Summing the two expressions results in the cancellation of odd-powered terms of $y$:

$(x + y)^4 + (x - y)^4 = 2[\binom{4}{0}x^4 + \binom{4}{2}x^2y^2 + \binom{4}{4}y^4]$

Calculating the binomial coefficients: $\binom{4}{0} = 1$, $\binom{4}{2} = 6$, $\binom{4}{4} = 1$.

The simplified expression is: $2[x^4 + 6x^2y^2 + y^4]$.

Step 3: Substitution of Original Variables

Substitute $x = a^2$ and $y^2 = a^2 - 1$ back into the simplified polynomial:

• $x^4 = (a^2)^4 = a^8$
• $x^2y^2 = (a^2)^2(a^2 - 1) = a^4(a^2 - 1) = a^6 - a^4$
• $y^4 = (a^2 - 1)^2 = a^4 - 2a^2 + 1$

Step 4: Final Algebraic Simplification

Substitute these values into the expression $2[x^4 + 6x^2y^2 + y^4]$:

$= 2[a^8 + 6(a^6 - a^4) + (a^4 - 2a^2 + 1)]$

$= 2[a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1]$

Combine the like terms inside the brackets:

$= 2[a^8 + 6a^6 - 5a^4 - 2a^2 + 1]$

Distributing the constant 2:

$= 2a^8 + 12a^6 - 10a^4 - 4a^2 + 2$

Visual Representation of Symmetric Expansion

Final Solution: 2a^{8} + 12a^{6} - 10a^{4} - 4a^{2} + 2

Step 1: Establishing the Binomial Expansion

To approximate $(0.99)^5$, we express the base as a binomial sum: $(1 - 0.01)^5$. We utilize the Binomial Theorem, which states that for any $n \in \mathbb{N}$:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Setting $a = 1$, $b = -0.01$, and $n = 5$, the expansion is:

$(1 - 0.01)^5 = \binom{5}{0}(1)^5(-0.01)^0 + \binom{5}{1}(1)^4(-0.01)^1 + \binom{5}{2}(1)^3(-0.01)^2 + \dots$ [Per the Binomial Theorem Expansion]

Step 2: Calculating the First Three Terms

We restrict the calculation to the first three terms as per the requirement:

• Term 1 ($k=0$): $\binom{5}{0} (1)^5 (1) = 1 \times 1 \times 1 = 1$
• Term 2 ($k=1$): $\binom{5}{1} (1)^4 (-0.01)^1 = 5 \times 1 \times -0.01 = -0.05$
• Term 3 ($k=2$): $\binom{5}{2} (1)^3 (-0.01)^2 = \frac{5 \times 4}{2 \times 1} \times 0.0001 = 10 \times 0.0001 = 0.001$

Step 3: Summation of the Terms

Summing these components yields the approximation:

$S \approx 1 - 0.05 + 0.001$ $S \approx 0.95 + 0.001$ $S \approx 0.951$

Geometric Representation of Binomial Coefficients

Final Solution: The approximation of $(0.99)^5$ using the first three terms is $0.951$.

Step 1: Theoretical Foundation via Binomial Expansion

To prove that $(a - b)$ is a factor of $(a^n - b^n)$, we utilize the Binomial Theorem, which states that for any positive integer $n$:

$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ [Per the Binomial Theorem Definition].

Step 2: Substitution and Expansion

Following the hint, we express $a$ as $(a - b + b)$. Thus, we consider the expression $a^n$ rewritten as:

$a^n = ((a - b) + b)^n$

Applying the Binomial Theorem by setting $x = (a - b)$ and $y = b$:

$a^n = \sum_{k=0}^{n} \binom{n}{k} (a - b)^{n-k} b^k$

$a^n = \binom{n}{0}(a - b)^n b^0 + \binom{n}{1}(a - b)^{n-1} b^1 + \dots + \binom{n}{n-1}(a - b)^1 b^{n-1} + \binom{n}{n}(a - b)^0 b^n$

Step 3: Algebraic Simplification

We observe that all terms in the summation contain at least one factor of $(a - b)$, except for the final term where $(a - b)^0 = 1$:

$a^n = [(a - b)^n + \binom{n}{1}(a - b)^{n-1} b + \dots + \binom{n}{n-1}(a - b) b^{n-1}] + b^n$

Factor out $(a - b)$ from all terms within the brackets:

$a^n = (a - b) \left[ (a - b)^{n-1} + \binom{n}{1}(a - b)^{n-2} b + \dots + \binom{n}{n-1} b^{n-1} \right] + b^n$

Step 4: Final Derivation of the Factor

Subtract $b^n$ from both sides of the equation:

$a^n - b^n = (a - b) \left[ \sum_{k=0}^{n-1} \binom{n}{k} (a - b)^{n-k-1} b^k \right]$

Since the expression within the square brackets is a sum of products of integers $(a, b, n)$, it results in an integer, let us denote it as $I$. Therefore:

$a^n - b^n = (a - b) \cdot I$

By the definition of divisibility, if $a^n - b^n$ can be expressed as the product of $(a - b)$ and an integer $I$, then $(a - b)$ must be a factor of $(a^n - b^n)$ [Per the Euclidean Property of Integers].

Visual Representation of Factorization Logic

Final Solution: Since $(a^n - b^n) = (a - b) \times I$, it is proven that $(a - b)$ is a factor of $(a^n - b^n)$ for all positive integers $n$.

Step 1: Application of the Binomial Theorem

According to the Binomial Theorem, the expansion of $(a+b)^n$ is given by:

$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Let $x = \sqrt{3}$ and $y = \sqrt{2}$. The given expression is of the form $(x+y)^6 - (x-y)^6$.

The expansion for $(x+y)^6$ is: $(x+y)^6 = \binom{6}{0}x^6 + \binom{6}{1}x^5y + \binom{6}{2}x^4y^2 + \binom{6}{3}x^3y^3 + \binom{6}{4}x^2y^4 + \binom{6}{5}xy^5 + \binom{6}{6}y^6$

The expansion for $(x-y)^6$ is: $(x-y)^6 = \binom{6}{0}x^6 - \binom{6}{1}x^5y + \binom{6}{2}x^4y^2 - \binom{6}{3}x^3y^3 + \binom{6}{4}x^2y^4 - \binom{6}{5}xy^5 + \binom{6}{6}y^6$

Step 2: Simplifying the Difference

Subtracting $(x-y)^6$ from $(x+y)^6$, the even-positioned terms (relative to the index $k$ starting at 0) cancel out, leaving twice the sum of the odd-indexed terms:

$(x+y)^6 - (x-y)^6 = 2 \left[ \binom{6}{1}x^5y + \binom{6}{3}x^3y^3 + \binom{6}{5}xy^5 \right]$ [By the property of binomial symmetry where $(a+b)^n - (a-b)^n = 2 \sum_{k \text{ odd}} \binom{n}{k} a^{n-k} b^k$]

Step 3: Calculating Binomial Coefficients

Combination	Calculation	Result
$\binom{6}{1}$	$\frac{6!}{1!5!}$	$6$
$\binom{6}{3}$	$\frac{6 \times 5 \times 4}{3 \times 2 \times 1}$	$20$
$\binom{6}{5}$	$\binom{6}{1}$	$6$

Step 4: Substitution and Evaluation

Substitute $x = \sqrt{3}$ and $y = \sqrt{2}$ into the simplified expression:

$= 2 \left[ 6(\sqrt{3})^5(\sqrt{2}) + 20(\sqrt{3})^3(\sqrt{2})^3 + 6(\sqrt{3})(\sqrt{2})^5 \right]$

Evaluate the powers:

• $(\sqrt{3})^5 = 9\sqrt{3}$; $(\sqrt{2}) = \sqrt{2} \implies 6(9\sqrt{3})(\sqrt{2}) = 54\sqrt{6}$
• $(\sqrt{3})^3 = 3\sqrt{3}$; $(\sqrt{2})^3 = 2\sqrt{2} \implies 20(3\sqrt{3})(2\sqrt{2}) = 120\sqrt{6}$
• $(\sqrt{3}) = \sqrt{3}$; $(\sqrt{2})^5 = 4\sqrt{2} \implies 6(\sqrt{3})(4\sqrt{2}) = 24\sqrt{6}$

Summing the terms inside the bracket: $= 2 [54\sqrt{6} + 120\sqrt{6} + 24\sqrt{6}]$ $= 2 [198\sqrt{6}]$ $= 396\sqrt{6}$

Step 5: Visual Representation

Final Solution: 396\sqrt{6}