Answer: C
Let \(a = 3 + 2\sqrt{2}\) Then, \[ \frac{1}{a} = \frac{1}{3 + 2\sqrt{2}} \] Rationalizing the denominator: Here’s the English translation of your solution: Let \(a = 3 + 2\sqrt{2}\) Then, \[ \frac{1}{a} = \frac{1}{3 + 2\sqrt{2}} \] Rationalizing the denominator: \[ = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} \] \[ = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} \] Therefore, \[ a + \frac{1}{a} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \] Therefore, \[ a + \frac{1}{a} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \] Now, \[ \frac{a^6 + a^4 + a^2 + 1}{a^3} = a^3 + a + \frac{1}{a} + \frac{1}{a^3} = a^3 + \frac{1}{a^3} + a + \frac{1}{a} \] Now, \[ \frac{a^6 + a^4 + a^2 + 1}{a^3} = a^3 + a + \frac{1}{a} + \frac{1}{a^3} = a^3 + \frac{1}{a^3} + a + \frac{1}{a} \] Using the identityUsing the identity: \[ a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3a\cdot\frac{1}{a}\left(a + \frac{1}{a}\right) = 6^3 - 3 \cdot 6 + 6 = 216 - 18 + 6 = 204 \] : \[ a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3a\cdot\frac{1}{a}\left(a + \frac{1}{a}\right) = 6^3 - 3 \cdot 6 + 6 = 216 - 18 + 6 = 204 \]
Let \(a = 3 + 2\sqrt{2}\) Then, \[ \frac{1}{a} = \frac{1}{3 + 2\sqrt{2}} \] Rationalizing the denominator: Here’s the English translation of your solution: Let \(a = 3 + 2\sqrt{2}\) Then, \[ \frac{1}{a} = \frac{1}{3 + 2\sqrt{2}} \] Rationalizing the denominator: \[ = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} \] \[ = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} \] Therefore, \[ a + \frac{1}{a} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \] Therefore, \[ a + \frac{1}{a} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \] Now, \[ \frac{a^6 + a^4 + a^2 + 1}{a^3} = a^3 + a + \frac{1}{a} + \frac{1}{a^3} = a^3 + \frac{1}{a^3} + a + \frac{1}{a} \] Now, \[ \frac{a^6 + a^4 + a^2 + 1}{a^3} = a^3 + a + \frac{1}{a} + \frac{1}{a^3} = a^3 + \frac{1}{a^3} + a + \frac{1}{a} \] Using the identityUsing the identity: \[ a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3a\cdot\frac{1}{a}\left(a + \frac{1}{a}\right) = 6^3 - 3 \cdot 6 + 6 = 216 - 18 + 6 = 204 \] : \[ a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3a\cdot\frac{1}{a}\left(a + \frac{1}{a}\right) = 6^3 - 3 \cdot 6 + 6 = 216 - 18 + 6 = 204 \]