Question

if a-1/a =8 and a is not equal to 0 find a^2-1/a^2.
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Answer 1

Answer:

Given,

$\sf \: a - \frac{1}{a} = 8 \\ \\ \sf \: \odot \: \: {a}^{2} - \frac{1}{ {a}^{2} } = \: to \: find \\ \\ \bf \: now \\ \\ \sf{(a + \frac{1}{a} })^{2} = {(a - \frac{1}{a} })^{2} + 4 \times a \times \frac{1}{a} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {8}^{2} + 4 \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 64 + 4 = 68 \\ \sf \implies \: a + \frac{1}{a} = \sqrt{68} \\ \\ \bf \: now \\ \\ \sf \: {a}^{2} - \frac{1}{ {a}^{2} } \\ \sf \: = (a + \frac{1}{a} )(a - \frac{1}{a} ) \\ \sf \: = \sqrt{68} \: \: \times \: 8 \\ \bf \: = 8 \sqrt{68} \\ \bf \: = 8 \times \sqrt{17 \times 4} \\ \bf \: = 8 \times 2 \times \sqrt{17} \\ \bf \: = 16 \sqrt{17}$

Additional Information :-

Important identities :-

$\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\boxed{ \bf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\boxed{ \bf{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}$

$\boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

$\boxed{ \bf{ \: {x}^{2} - {y}^{2} = (x - y)(x + y)}}$

$\boxed{ \bf{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}}$

$\boxed{ \bf{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} )}}$

$\boxed{ \bf{ \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2} )}}$

Answer 2

Answer:

$Given,</p><p>\begin{gathered} \sf \: a - \frac{1}{a} = 8 \\ \\ \sf \: \odot \: \: {a}^{2} - \frac{1}{ {a}^{2} } = \: to \: find \\ \\ \bf \: now \\ \\ \sf{(a + \frac{1}{a} })^{2} = {(a - \frac{1}{a} })^{2} + 4 \times a \times \frac{1}{a} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {8}^{2} + 4 \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 64 + 4 = 68 \\ \sf \implies \: a + \frac{1}{a} = \sqrt{68} \\ \\ \bf \: now \\ \\ \sf \: {a}^{2} - \frac{1}{ {a}^{2} } \\ \sf \: = (a + \frac{1}{a} )(a - \frac{1}{a} ) \\ \sf \: = \sqrt{68} \: \: \times \: 8 \\ \bf \: = 8 \sqrt{68} \\ \bf \: = 8 \times \sqrt{17 \times 4} \\ \bf \: = 8 \times 2 \times \sqrt{17} \\ \bf \: = 16 \sqrt{17} \end{gathered}a−a1=8⊙a2−a21=tofindnow(a+a1)2=(a−a1)2+4×a×a1=82+4=64+4=68⟹a+a1=68nowa2−a21=(a+a1)(a−a1)=68×8=868=8×17×4=8×2×17=1617</p><p>Additional Information :-</p><p>Important identities :-</p><p>\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}(x+y)2=x2+y2+2xy</p><p>\boxed{ \bf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}(x−y)2=x2+y2−2xy</p><p>\boxed{ \bf{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}(x−y)3=x3−y3−3xy(x−y)</p><p>\boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}(x+y)3=x3+y3+3xy(x+y)</p><p>\boxed{ \bf{ \: {x}^{2} - {y}^{2} = (x - y)(x + y)}}x2−y2=(x−y)(x+y)</p><p>\boxed{ \bf{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}}x3+y3=(x+y)(x2−xy+y2)</p><p>\boxed{ \bf{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} )}}x3−y3=(x−y)(x2+xy+y2)</p><p>\boxed{ \bf{ \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2} )}}x4−y4=(x−y)(x+y)(x2+y2)</p><p>$