Question

if a minus one upon a is equal to 8 find a+ 1 upon a and a square minus one upon a square please solve it quickly and do it with expansions method​

Answer 1

Step-by-step explanation:

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Answer 2

$a - \frac{1}{a} = 8 \\ squaring \: on \: both \: sides \\ {a }^{2} + \frac{1}{a ^{2} } - 2 = 64 \\ {a^{2} + \frac{1}{a^{2} } } = 66 \\ now \\ {a}^{2} + \frac{1}{ {a}^{2} } + 2 = 66 + 2 \\ (a + \frac{1}{a} ) ^{2} = 68 \\ a + \frac{1}{a} = ± \sqrt{68} \\a + \frac{1}{a } = ±2 \sqrt{17} \\ now \\ {a }^{2} - \frac{1}{ {a}^{2} } =( a + \frac{1}{a} )(a - \frac{1}{a} ) \\ {a }^{2} - \frac{1}{ {a}^{2} } =(± 2\sqrt{17} )(8) =±16 \sqrt{17}$