Question

If a⁴+1\a⁴=727, then the value of a³—1\a³ is ________.

Answer 1

Here's your answer!!

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It is given that,

$= > {a}^{4} + \frac{1}{ {a}^{4} } = 727$

By adding 2 both side ,we get :-

$= > {a}^{4} + \frac{1}{ {a}^{4} } + 2 = 727 + 2$

We can write it as ,

$= > {( {a}^{2} )}^{2} + \frac{1}{ { ({a}^{2}) }^{2} } + 2 \times {a}^{2} \times \frac{1}{ {a}^{2} } = 729$

$= > {( {a}^{2} + \frac{1}{ {a}^{2} } )}^{2} = {(27)}^{2}$

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$identity \: used = \\ = > \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2 \times a \times b)$
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$= > {a}^{2} + \frac{1}{ {a}^{2} } = 27$

Now,

By subtracting 2 from both side ,we get :-

$= > {a}^{2} + \frac{1}{ {a}^{2} } - 2 = 27 - 2$

We can write it as :-

$= > {a}^{2} + \frac{1}{ {a}^{2} } - 2 \times a \times \frac{1}{ a} = 27 - 2$

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$identity \: used = \: \\ = > {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2 \times a \times b$
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$= > (a - \frac{1}{a} ) = 5.......(i)$

Now,

By cubing both side , we get :-

$= > {(a + \frac{1}{a} )}^{3} = {(5)}^{3}$

$= > {a}^{3} - \frac{1}{ {a}^{3} } - 3 \times a \times \frac{1}{a} (a - \frac{1}{a} ) = 125$

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$identity \: used = \\ = > {(a - b)}^{3} = {a}^{3} - {b}^{3} - 3ab(a - b)$
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By substituting value of equation (i) ,we get :-

$= > {a}^{3} - \frac{1}{ {a}^{3} } - 3 \times a \times \frac{1}{a} (5) = 125$

$= > ({a}^{3} - \frac{1}{ {a}^{3} } ) - 15 = 125$

$= > ({a}^{3} - \frac{1}{ {a}^{3} } ) = 125 + 15$

$= > ( {a}^{3} - \frac{1}{ {a}^{3} }) = 140$

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Hope it helps you!! :)

Answer 2

Answer:

140

Step-by-step explanation:

I think it's right answer