If \( x=(2+\sqrt{3})^{\frac{1}{3}}+(2+\sqrt{3})^{-\frac{1}{3}} \) and \(x^{3}-3x+k=0\), then the value of k is:
Step-by-step Solution:
Question: If \( x=(2+\sqrt{3})^{\frac{1}{3}}+(2+\sqrt{3})^{-\frac{1}{3}} \) and \(x^{3}-3x+k=0\), then the value of \(k\) is:
Let \( a = (2+\sqrt{3})^{1/3} \). Then the given expression becomes:
\[ x = a + \frac{1}{a} \]
We use the identity:
\[ (a + \frac{1}{a})^3 = a^3 + \frac{1}{a^3} + 3(a + \frac{1}{a}) \]
Here, \( a^3 = 2+\sqrt{3} \) and \( \frac{1}{a^3} = \frac{1}{2+\sqrt{3}} \). Let's compute \( a^3 + \frac{1}{a^3} \):
\[ \frac{1}{2+\sqrt{3}} \cdot \frac{2-\sqrt{3}}{2-\sqrt{3}} = 2-\sqrt{3} \]
\[ a^3 + \frac{1}{a^3} = (2+\sqrt{3}) + (2-\sqrt{3}) = 4 \]
So, using the identity:
\[ x^3 = a^3 + \frac{1}{a^3} + 3x = 4 + 3x \]
Rewriting in the form of the given equation \( x^3 - 3x + k = 0 \):
\[ x^3 - 3x + k = 0 \implies 4 + 3x - 3x + k = 0 \implies k + 4 = 0 \implies k = -4 \]
Answer: A. -4