Binomial Theorem

To find the degree of the polynomial, we expand the expression completely using the Binomial Theorem. This question contains a deliberate trap: the highest power ($$x^{12}$$) completely cancels out during simplification.

Step 1: Apply the Binomial Theorem expansion
Let $$a = x^2 - 3$$ and $$b = (1 - x^4)^{\frac{1}{2}}$$. The expression is in the form $$(a + b)^6 + (a - b)^6$$.

Expanding both parts with the Binomial Theorem and adding them together cancels out all terms with odd powers of $$b$$:
$$(a+b)^6 + (a-b)^6 = 2 \left[ \binom{6}{0}a^6 + \binom{6}{2}a^4b^2 + \binom{6}{4}a^2b^4 + \binom{6}{6}b^6 \right]$$

Substitute the combination values $$\binom{6}{0}=1$$, $$\binom{6}{2}=15$$, $$\binom{6}{4}=15$$, and $$\binom{6}{6}=1$$:
$$= 2 [ a^6 + 15a^4b^2 + 15a^2b^4 + b^6 ]$$

Step 2: Write out the full expansion of each component
Substitute $$a = x^2 - 3$$ and $$b^2 = 1 - x^4$$ back into the formula:
$$= 2 [ (x^2 - 3)^6 + 15(x^2 - 3)^4(1 - x^4) + 15(x^2 - 3)^2(1 - x^4)^2 + (1 - x^4)^3 ]$$

We fully expand each of the four components inside the brackets:

1. For $$(x^2 - 3)^6$$:
$$x^{12} - 18x^{10} + 135x^8 - 540x^6 + 1215x^4 - 1458x^2 + 729$$

2. For $$15(x^2 - 3)^4(1 - x^4)$$:
$$15(x^8 - 12x^6 + 54x^4 - 108x^2 + 81)(1 - x^4)$$
$$= -15x^{12} + 15x^8 + 180x^{10} - 180x^6 - 810x^8 + 810x^4 + 1620x^6 - 1620x^2 - 1215x^4 + 1215$$
$$= -15x^{12} + 180x^{10} - 795x^8 + 1440x^6 - 405x^4 - 1620x^2 + 1215$$

3. For $$15(x^2 - 3)^2(1 - x^4)^2$$:
$$15(x^4 - 6x^2 + 9)(x^8 - 2x^4 + 1)$$
$$= 15x^{12} - 30x^8 + 15x^4 - 90x^{10} + 180x^6 - 90x^2 + 135x^8 - 270x^4 + 135$$
$$= 15x^{12} - 90x^{10} + 105x^8 + 180x^6 - 255x^4 - 90x^2 + 135$$

4. For $$(1 - x^4)^3$$:
$$-x^{12} + 3x^8 - 3x^4 + 1$$

Step 3: Combine all terms to get the complete polynomial expression
Now, sum all four fully expanded parts together inside the brackets:
$$\text{Sum} = (x^{12} - 15x^{12} + 15x^{12} - x^{12}) + (-18x^{10} + 180x^{10} - 90x^{10}) + (135x^8 - 795x^8 + 105x^8 + 3x^8) + (-540x^6 + 1440x^6 + 180x^6) + (1215x^4 - 405x^4 - 255x^4 - 3x^4) + (-1458x^2 - 1620x^2 - 90x^2) + (729 + 1215 + 135 + 1)$$

Simplify each grouped power step-by-step:
* For $$x^{12}$$: $$1 - 15 + 15 - 1 = 0$$
* For $$x^{10}$$: $$-18 + 180 - 90 = 72$$
* For $$x^8$$: $$135 - 795 + 105 + 3 = -552$$
* For $$x^6$$: $$-540 + 1440 + 180 = 1080$$
* For $$x^4$$: $$1215 - 405 - 255 - 3 = 552$$
* For $$x^2$$: $$-1458 - 1620 - 90 = -3168$$
* For constants: $$729 + 1215 + 135 + 1 = 2080$$

Multiplying the entire simplified bracket by the front factor of $$2$$ gives the complete expanded polynomial:
$$\text{Complete Expression} = 2(72x^{10} - 552x^8 + 1080x^6 + 552x^4 - 3168x^2 + 2080)$$
$$\text{Complete Expression} = 144x^{10} - 1104x^8 + 2160x^6 + 1104x^4 - 6336x^2 + 4160$$

Since the highest remaining power of $$x$$ is 10, the true degree of the polynomial is 10.

To find the fractional part of the expression, we need to find the remainder when $$3^{2026}$$ is divided by 80.

We rewrite the base of the exponent to match a number close to a multiple of 80 by noticing that $$3^4 = 81 = 80 + 1$$. 

First, we split the exponent to express it in terms of a power of 4 by writing $$2026 = 4 \times 506 + 2$$. 

This allows us to separate the terms into $$3^{2026} = 3^2 \times (3^4)^{506} = 9 \times (81)^{506}$$. 

Substituting $$81 = 80 + 1$$ into the expression yields $$9 \times (80 + 1)^{506}$$. 

Applying the Binomial Theorem to expand the two-term expression yields $$9 \times \left[ \binom{506}{0}(80)^{506} + \binom{506}{1}(80)^{505} + \dots + \binom{506}{505}(80)^1 + \binom{506}{506}(1)^{506} \right]$$. 

Every term inside the brackets except the last one contains a factor of 80, so we can group them together as an integer multiple $$80M$$. This simplifies our expression to $$9 \times (80M + 1) = 720M + 9$$. 

Dividing this entire expression by 80 gives $$\frac{720M + 9}{80} = 9M + \frac{9}{80}$$. Since $$9M$$ is an integer, the fractional part of the number is exactly $$\frac{9}{80}$$. 

Comparing this to the given form $$\frac{k}{80}$$ tells us that $$k = 9$$. The integer 9 can be factored into prime numbers as $$3^2$$, which means it has only 1 distinct prime factor, namely 3.

Hence, the final answer is 1

1. Factorise the First Base Expression

We begin by simplifying the first expression inside the exponent using grouping:

$$1 + x + y + \frac{x}{y} = (1 + x) + y\left(1 + \frac{1}{y}\right)$$

This layout is easier to handle if rearranged differently to isolate common binomials:

$$1 + \frac{x}{y} + x + y = \frac{y + x}{y} + (x + y)$$

Factoring out the shared numerator $$(x + y)$$:

$$(x + y)\left(\frac{1}{y} + 1\right) = (x + y)\left(1 + \frac{1}{y}\right)$$

Substituting this product back into the power of $$10$$ gives the first simplified component:

$$f_1(x, y) = (x + y)^{10} \left(1 + \frac{1}{y}\right)^{10}$$

2. Factorise the Second Base Expression

Now we apply a similar grouping approach to the second expression:

$$1 + y + \frac{1}{x} + \frac{y}{x} = (1 + y) + \frac{1}{x}(1 + y)$$

Factoring out the shared expression $$(1 + y)$$:

$$(1 + y)\left(1 + \frac{1}{x}\right)$$

Substituting this back into the power of $$10$$ gives the second simplified component:

$$f_2(x, y) = (1 + y)^{10} \left(1 + \frac{1}{x}\right)^{10}$$

3. Determine the Unique Exponent Pairs for Each Component

We analyze the exponent footprints $$(a, b)$$ of the terms generated by each expanded part independently.

For the first component:

$$f_1(x, y) = \left( \sum_{i=0}^{10} \binom{10}{i} x^{10-i} y^i \right) \left( \sum_{j=0}^{10} \binom{10}{j} y^{-j} \right) = \sum_{i=0}^{10} \sum_{j=0}^{10} \binom{10}{i} \binom{10}{j} x^{10-i} y^{i-j}$$

The exponent pairs generated are:

$$a = 10 - i \quad \text{and} \quad b = i - j$$

Since $$0 \leq i \leq 10$$ and $$0 \leq j \leq 10$$, each choice of the pair $$(i, j)$$ yields a unique set of exponents. The total number of distinct terms from the first component is:

$$11 \times 11 = 121 \text{ terms}$$

For the second component:

$$f_2(x, y) = \left( \sum_{k=0}^{10} \binom{10}{k} y^k \right) \left( \sum_{m=0}^{10} \binom{10}{m} x^{-m} \right) = \sum_{k=0}^{10} \sum_{m=0}^{10} \binom{10}{k} \binom{10}{m} x^{-m} y^k$$

The exponent pairs generated are:

$$a = -m \quad \text{and} \quad b = k$$

Since $$0 \leq k \leq 10$$ and $$0 \leq m \leq 10$$, every pair $$(k, m)$$ produces a unique set of exponents. The total number of distinct terms from the second component is:

$$11 \times 11 = 121 \text{ terms}$$

4. Find and Subtract the Overlapping Terms

When combining the two expansions, terms with identical exponent pairs $$(a, b)$$ will merge together. To find the true number of unique terms, we use the Principle of Inclusion-Exclusion:

$$\text{Total Terms } T = \text{Terms}(f_1) + \text{Terms}(f_2) - \text{Terms}(f_1 \cap f_2)$$

Let's find when an exponent pair $$(a, b)$$ belongs to both expansions simultaneously:

From the first expansion: $$a = 10 - i \geq 0$$ (since $$i \leq 10$$).

From the second expansion: $$a = -m \leq 0$$ (since $$m \geq 0$$).

For a term to exist in both, its $$x$$-exponent $$a$$ must satisfy both conditions:

$$a \geq 0 \quad \text{and} \quad a \leq 0 \implies a = 0$$

Setting $$a = 0$$ determines the exact matching indices for both expansions:

- In the first expansion: $$10 - i = 0 \implies i = 10$$.

- In the second expansion: $$-m = 0 \implies m = 0$$.

Now we analyze the corresponding $$y$$-exponent $$b$$ under these index constraints:

- From the first expansion with $$i = 10$$: $$b = 10 - j$$. Since $$0 \leq j \leq 10$$, this produces $$b \in \{0, 1, 2, \dots, 10\}$$.

- From the second expansion with $$m = 0$$: $$b = k$$. Since $$0 \leq k \leq 10$$, this also produces $$b \in \{0, 1, 2, \dots, 10\}$$.

The overlapping pairs are precisely the $$11$$ distinct terms where $$a = 0$$ and $$0 \leq b \leq 10$$:

$$\text{Overlapping Terms} = 11$$

5. Calculate the Final Value

Applying these values to our inclusion-exclusion formula:

$$T = 121 + 121 - 11$$

$$T = 242 - 11 = 231$$