Answer :
Answer:
\[
x^3 + \frac{1}{x^3} = 198
\]
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Explanation:
To find the value of \( x^3 + \frac{1}{x^3} \) given \( x = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \), we start by simplifying \( x \).
First, rationalize the denominator of \( x \):
\[
x = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)(\sqrt{2} + 1)}
\]
Simplify the denominator:
\[
(\sqrt{2} - 1)(\sqrt{2} + 1) = 2 - 1 = 1
\]
Simplify the numerator:
\[
(\sqrt{2} + 1)^2 = (\sqrt{2})^2 + 2 \cdot \sqrt{2} \cdot 1 + 1^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}
\]
Thus,
\[
x = 3 + 2\sqrt{2}
\]
Next, find \( \frac{1}{x} \):
\[
\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}}
\]
Rationalize the denominator:
\[
\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} \cdot \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})}
\]
Simplify the denominator:
\[
(3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 9 - (2\sqrt{2})^2 = 9 - 8 = 1
\]
So,
\[
\frac{1}{x} = 3 - 2\sqrt{2}
\]
Next, calculate \( x + \frac{1}{x} \):
\[
x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6
\]
We need \( x^3 + \frac{1}{x^3} \). Using the identity for cubes,
\[
x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3 \left( x + \frac{1}{x} \right)
\]
Let \( y = x + \frac{1}{x} \). Then,
\[
x^3 + \frac{1}{x^3} = y^3 - 3y
\]
Substitute \( y = 6 \):
\[
x^3 + \frac{1}{x^3} = 6^3 - 3 \cdot 6 = 216 - 18 = 198
\]
Therefore,
\[
x^3 + \frac{1}{x^3} = 198
\]