Question 69

If $$a + \frac{1}{a} = 3$$, then the value of $$\left(a^6 + \frac{1}{a^6}\right)$$ is equal to:

Solution

$$a + \frac{1}{a} = 3$$,

Cubing both side,

$$ \left(a + \frac{1}{a}\right)^{3}=3^{3}$$

$$\left(a^{3} + \frac{1}{a^{3}}\right)+9=27$$

$$a^{3} + \frac{1}{a^{3}}=27-9=18$$

Again squaring both side,

$$\left(a^{3} + \frac{1}{a^{3}}\right)^{2}=18^{2}$$

$$\left(a^{6} + \frac{1}{a^{6}}\right)+2=324$$

$$\left(a^{6} + \frac{1}{a^{6}}\right)=324-2=322$$

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