Find k given \(x=(2+\sqrt{3})^{1/3}+(2+\sqrt{3})^{-1/3}\) and \(x^{3}-3x+k=0\)

1. Let \(a=2+\sqrt{3}\Rightarrow a^{-1}=2-\sqrt{3}\). Define \(u^3=a,\;v^3=a^{-1}\) so that \(uv=\big(a\cdot a^{-1}\big)^{1/3}=1\) and \(x=u+v\).
2. Use \((u+v)^3=u^3+v^3+3uv(u+v)\):
   \[ x^3=u^3+v^3+3uv(u+v)=(2+\sqrt{3})+(2-\sqrt{3})+3x=4+3x. \]
3. Rearrange: \(\;x^3-3x-4=0\). Comparing with \(x^{3}-3x+k=0\) gives \(\;k=-4\).

Given: Polynomial $$f(x) = 2x^2 - 5x + k$$ and $(x-1)$ is a factor.

Step 1: Apply factor theorem.
If $(x-1)$ is a factor, then $f(1) = 0$.

Step 2: Substitute $x=1$.
$$f(1) = 2(1)^2 - 5(1) + k = 2 - 5 + k = -3 + k$$

Step 3: Solve for $k$.
$$-3 + k = 0 \quad \Rightarrow \quad k = 3$$

Value of k: 3

Answer: Option 3

Given \(x^2+\dfrac{1}{x^2}=2\). Let \(t=x+\dfrac{1}{x}\). Then \(x^2+\dfrac{1}{x^2}=t^2-2\), so \(t^2-2=2 \Rightarrow t^2=4 \Rightarrow t=\pm2\).

From \(x+\dfrac{1}{x}=2 \Rightarrow x=1\) and from \(x+\dfrac{1}{x}=-2 \Rightarrow x=-1\).

Hence \(x^{256}+\dfrac{1}{x^{256}}= \begin{cases} 1+1=2,& x=1\\ (-1)^{256}+(-1)^{-256}=1+1=2,& x=-1 \end{cases}\).

Final Answer: \(2\).