Question

x + 1/x =3 find the value of x3 + 1/x3

Answer 1

Hi friend
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Given that : -
$\boxed{x + \frac{1}{x} = 3}$

To find : -
$x {}^{3} + \frac{1}{x {}^{3} } = ?$

Now,

x + 1/x = 3

On squaring both sides,

(x + 1/x)² = (3)²

=> x² + 2 × x × 1/x + 1/x² = 9

=> x² + 1/x² = 9 - 2

=> x² + 1/x² = 7 ......(i)

Then,

x³ + 1/x³

= (x + 1/x)(x² - x ×1/x + 1/x²). [using identity : a³ + b³ = (a + b)(a² - ab + b²)]

= 3 × (x² + 1/x² - 1)

= 3 × (7 - 1). [From (i)]

= 3 × 6

= 18

HOPE IT HELPS

Answer 2

Answer:

$Using \: \: the \: \: identity : \\ {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b) \\ Here, \\ a = x \\ b = \frac{1}{x}$

Step-by-step explanation:

$x + \frac{1}{x} = 3 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = {3}^{3} \\ {x}^{3} + \frac{1}{ {x}^{3} } + (3 \times \cancel x \times \frac{1}{ \cancel x} )(x + \frac{1}{x} ) = 27 \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(3) = 27 \\ {x}^{3} + \frac{1}{ {x}^{3} } + 9 = 27 \\ {x}^{3} + \frac{1}{ {x}^{3} } = 27 - 9 \\ \boxed{ \large {x}^{3} + \frac{1}{ {x}^{3} } = \underline{ \underline{ 18}}}$