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If a+1/a=6 find a- 1/a and a^2- 1/a^2

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GIVEN THAT,

a + \frac{1}{a} = 6 \\ \\ Squaring \: both \: sides \\ we \:get \\ \\ {(a + \frac{1}{a} ) }^{2} = {6}^{2} \\ \\ {a}^{2} + { \frac{1}{a {}^{2} } } + 2 \times a \times \frac{1}{a} = 36 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } + 2 = 36 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } = 34 \\ \\ Subtracting \: \: 2\: \: both \: \: side s \\ we \: get \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } - 2 = 34 - 2 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } - 2 \times a \times \frac{1}{a} = 32 \\ \\ {(a - \frac{1}{a} ) }^{2} = 32 \\ \\ {(a - \frac{1}{a} ) }= \sqrt{32} \:\: or \: \: -\sqrt{32} \\ \\ {(a - \frac{1}{a} ) }= 4\sqrt{2} \: \: or \: \: - 4 \sqrt{2} \\ \\Now, \\\\ {a}^{2} - \frac{1}{ {a}^{2} } = (a + \frac{1}{a} )(a - \frac{1}{a} ) \\ \\ when \: (a - \frac{1}{a} ) = 4 \sqrt{2 } \\ \\ (a {}^{2} - \frac{1}{a {}^{2} } ) = 6 \times 4 \sqrt{2} \\ \\ = 24 \sqrt{2} \\ \\ when \: (a - \frac{1}{a} ) = - 4 \sqrt{2} \\ \\ (a {}^{2} - \frac{1}{a {}^{2} } ) = 6 \times ( - 4 \sqrt{2} ) \\ \\ = - 24 \sqrt{2}

HENCE,

(a - \frac{1}{a} ) = 4 \sqrt{2} \: \: or \: \: - 4 \sqrt{2} \\ \\ and \\ \\ (a {}^{2} - \frac{1}{a {}^{2} } ) = 24 \sqrt{2} \: \: or \: \: - 24 \sqrt{2}


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