If $$\alpha$$, $$\beta \neq 0$$, $$f(n) = \alpha^n + \beta^n$$ and $$\begin{vmatrix} 3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4) \end{vmatrix} = K(1-\alpha)^2(1-\beta)^2(\alpha - \beta)^2$$, then K is equal to:

We have a function defined by $$f(n)=\alpha^{\,n}+\beta^{\,n}\quad$$ with $$\quad \alpha\neq 0,\;\beta\neq 0.$$ The given determinant is

$$ \Delta=\begin{vmatrix} 3 & 1+f(1) & 1+f(2)\\[4pt] 1+f(1) & 1+f(2) & 1+f(3)\\[4pt] 1+f(2) & 1+f(3) & 1+f(4) \end{vmatrix}. $$

According to the statement, this determinant can be factorised as

$$ \Delta=K\,(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}, $$

where $$K$$ is a constant independent of $$\alpha$$ and $$\beta$$. Because $$K$$ does not depend on the particular values of $$\alpha$$ and $$\beta$$ (as long as the factors on the right remain finite and non-zero), we can choose any convenient non-zero numbers for $$\alpha$$ and $$\beta$$ to evaluate both sides and isolate $$K$$. We only have to be careful to avoid the special cases that make the right-hand side vanish, i.e. we must not choose $$\alpha=1$$, $$\beta=1$$ or $$\alpha=\beta$$. A simple choice that keeps the arithmetic easy is

$$ \alpha=2,\qquad \beta=3. $$

With this choice we first compute all the required $$f(n)$$ values:

$$\begin{aligned} f(1)&=\alpha+\beta=2+3=5$$, $$\\ f(2)&=\alpha^{2}+\beta^{2}=4+9=13$$, $$\\ f(3)&=\alpha^{3}+\beta^{3}=8+27=35$$, $$\\ f(4)&=\alpha^{4}+\beta^{4}=16+81=97. \end{aligned}$$

Hence

$$1+f(1)=1+5=6$$, $$\qquad 1+f(2)=1+13=14$$, $$\qquad 1+f(3)=1+35=36$$, $$\qquad 1+f(4)=1+97=98.$$

Substituting these numerical values, the determinant becomes

$$ \Delta= \begin{vmatrix} 3 & 6 & 14\\[4pt] 6 & 14 & 36\\[4pt] 14 & 36 & 98 \end{vmatrix}. $$

We now evaluate this $$3\times3$$ determinant. The general formula for a determinant of order three is

$$ \begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix} =a(ei-hf)-b(di-gf)+c(dh-eg). $$

Identifying $$a=3$$, $$\; b=6$$, $$\; c=14$$, $$\; d=6$$, $$\; e=14$$, $$\; f=36$$, $$\; g=14$$, $$\; h=36$$, $$\; i=98$$ we obtain

$$ \begin{aligned} \Delta &=3\bigl(14\cdot98-36\cdot36\bigr) -6\bigl(6\cdot98-36\cdot14\bigr) +14\bigl(6\cdot36-14\cdot14\bigr).\\[6pt] \end{aligned} $$

We now carry out the arithmetic step by step:

$$\begin{aligned} 14\cdot98 &= 1372$$, $$\\ 36\cdot36 &= 1296$$, $$\\[4pt] 6\cdot98 &= 588$$, $$\\ 36\cdot14 &= 504$$, $$\\[4pt] 6\cdot36 &= 216$$, $$\\ 14\cdot14 &= 196. \end{aligned}$$

Substituting these into the expression for $$\Delta$$ we get

$$ \begin{aligned} \Delta &=3\bigl(1372-1296\bigr) -6\bigl(588-504\bigr) +14\bigl(216-196\bigr)\\[6pt] &=3\cdot76-6\cdot84+14\cdot20\\[6pt] &=228-504+280\\[6pt] &=508-504\\[6pt] &=4. \end{aligned} $$

Thus the numerical value of the determinant for our chosen $$\alpha$$ and $$\beta$$ is

$$ \boxed{\Delta=4}. $$

Next we compute the right-hand side factor

$$ (1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2} $$

for the same $$\alpha=2$$ and $$\beta=3$$:

$$ \begin{aligned} 1-\alpha &= 1-2 = -1 \quad\Longrightarrow\quad (1-\alpha)^{2}=(-1)^{2}=1,\\ 1-\beta &= 1-3 = -2 \quad\Longrightarrow\quad (1-\beta)^{2}=(-2)^{2}=4,\\ \alpha-\beta &= 2-3 = -1 \quad\Longrightarrow\quad (\alpha-\beta)^{2}=(-1)^{2}=1. \end{aligned} $$

Multiplying these three squares we find

$$ (1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}=1\cdot4\cdot1=4. $$

By the problem statement, for these same $$\alpha$$ and $$\beta$$ we must have

$$ \Delta = K\,(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}. $$

Substituting the numerical values we have just obtained, this gives

$$ 4 = K \times 4. $$

Dividing both sides by $$4$$ we immediately obtain

$$ K=1. $$

Since the choice of $$\alpha$$ and $$\beta$$ was arbitrary (aside from the obvious restrictions), this value of $$K$$ is valid for all permissible $$\alpha$$ and $$\beta$$. Hence, the factor multiplying $$(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$$ is $$1$$.

Hence, the correct answer is Option 1.