Question

if 3x-1/3x =5, find 81x⁴+1/81x⁴
​

Answer 1

$\tt {\underline{ \underline{ \purple{✠Answer:-}}}}$

Formula used:

$\sf{(a - \frac{1}{a} )^{2} = {a}^{2} + { (\frac{1}{a} )}^{2} - 2}$

Solution :

$\sf{(3x + \frac{1}{3})^{2} = (3x) ^{2} + \frac{1}{( {3x})^{2} } - 2 }$

$⟹ \sf{( {5})^{2} = 9 {x}^{2} + \frac{1}{9 {x}^{2} } - 2 }$

$⟹ \sf{25 + 2 = 9 {x}^{2} + \frac{1}{ {9x}^{2} } }$

$⟹ \sf{27 = 9 {x}^{2} + \frac{1}{ {9x}^{2} } }$

Now,

$\tt{(9 {x}^{2} + \frac{1}{9 {x}^{2} } ) ^{2} = (9x^{2})^{2} + (\frac{1}{9 {x}^{2} } )^{2} + 2 }$

$⟹ \tt{(27) ^{2} = 81 {x}^{4} + \frac{1}{81 {x}^{4} } + 2}$

$⟹ \tt{729 - 2 = 81 {x}^{4} + \frac{1}{81 {x}^{4} } }$

$⟹ \bf \red{727 = 81 {x}^{4} + \frac{1}{81 {x}^{4} } }$

Answer 2

Step-by-step explanation:

$\bf \underline{★Given-}$

$\rm{3x - \frac{1}{3x} = 5}$

$\bf \underline{★To \: find-}$

$\rm{the \: value \: of \: :81 {x}^{4} + \frac{1}{81 {x}^{4} } = \: ? }$

$\bf \underline{★Solution-}$

$\textsf{We have,}$

$\rm{3x - \frac{1}{3x} = 5}$

$\textsf{Squaring on both sides, we get}$

$\rm{ \bigg(3x - \frac{1}{3x} \bigg) ^{2} = (5 {)}^{2} }$

$\textsf{Now, comparing the given expression with (a-b)², we get}$

$\: \: \: \: \: \rm{a = 3x \: and \: b = \frac{1}{3x} }$

$\textsf{Using bionomial identity (a-b)² = a²-2ab+b², we have}$

$\rm{ \bigg(3x - \frac{1}{3x} \bigg) ^{2} = (5 {)}^{2} }$

$\rm{\implies \: (3x {)}^{2} - 2(3x) \bigg( \frac{1}{3x} \bigg) + \bigg(\frac{1}{3x} \bigg)^{2} = 25 }$

$\rm{\implies \:9x^{2} - 2( \cancel{3x}) \bigg( \frac{1}{ \cancel{3x}} \bigg) + \frac{1}{9 {x}^{2} } = 25}$

$\rm{\implies \:9 {x}^{2} - 2 + \frac{1}{9x^{2} } = 25}$

$\rm{\implies \:9 {x}^{2} + \frac{1}{9 {x}^{2} } = 25 + 2 }$

$\rm{\implies \:9 {x}^{2} + \frac{1}{9 {x}^{2} } = 27 }$

$\textsf{★Again, squaring on both sides, we get}$

.

$\rm{ \bigg(9 {x}^{2} + \frac{1}{9 {x}^{2} } \bigg)^{2} = (27 {)}^{2} }$

$\textsf{Now, comparing the given expression with (a+b)², we get}$

$\: \: \: \: \: \: \rm{a = 9 {x}^{2} \: and \: b = \frac{1}{9 {x}^{2} } }$

$\textsf{Using bionomial identity (a+b)² = a²+2ab+b², we have}$

$\rm{ \bigg(9 {x}^{2} + \frac{1}{9 {x}^{2} } \bigg)^{2} = (27 {)}^{2} }$

$\rm{ \implies(9 {x}^{2} {)}^{2} + 2(9 {x}^{2}) \bigg( \frac{1}{9 {x}^{2} } \bigg) + \bigg( \frac{1}{9 {x}^{2} } \bigg )^{2} } = (27 {)}^{2}$

$\rm{ \implies81 {x}^{4} + 2( \cancel{9 {x}^{2}} ) \bigg( \frac{1}{ \cancel{9 {x}^{2}} } \bigg) + \frac{1}{81 {x}^{4} } = 729}$

$\rm{ \implies81 {x}^{4} + 2 + \frac{1}{81 {x}^{4} } = 729}$

$\rm{ \implies81 {x}^{4} + \frac{1}{81 {x}^{4} } = 729 - 2}$

$\rm{ \implies81 {x}^{4} + \frac{1}{81 {x}^{4} } = 727 }$

$\bf{Hence, the \: value \: of \: : 81 {x}^{4} + \frac{1}{81 {x}^{4} } \: is \: 727.}$