Question

If x = 3-√8 ,find the value of x3+1/x3

Answer 1

Hey there !!


Given :-

$\bf x = 3 - \sqrt{8} .$

To find :-

$\bf {x}^{3} + \frac{1}{ {x}^{3} } .$


▶ Solution :-

$x = 3 - \sqrt{8} . \\ \\ \frac{1}{x} = \frac{1}{3 - \sqrt{8} } . \\ \\ x + \frac{1}{x} = 3 - \sqrt{8} + \frac{1}{3 - \sqrt{8} } . \\ \\ = \frac{ {(3 - \sqrt{8} )}^{2} + 1 }{3 - \sqrt{8} } . \\ \\ = \frac{ {3}^{2} + { (\sqrt{8}) }^{2} - 2 \times 3 \times \sqrt{8} + 1}{3 - \sqrt{8} } . \\ \\ = \frac{9 + 8 - 6 \sqrt{8} + 1}{3 - \sqrt{8} } . \\ \\ = \frac{18 - 6 \sqrt{8} }{3 - \sqrt{8} } \times \frac{3 + \sqrt{8} }{3 + \sqrt{8} } . \\ \\ = \frac{(18 - 6 \sqrt{8} )(3 + \sqrt{8} )}{ {3}^{2} - {( \sqrt{8} )}^{2} } . \\ \\ = \frac{54 + \cancel{18 \sqrt{8}} - \cancel{18 \sqrt{8} }- 48 }{9 - 8} . \\ \\ = 6. \\ \\ \\ \therefore x + \frac{1}{x} = 6. \\ (cubic \: both \: side) \\ \\ = > {(x + \frac{1}{x} )}^{3} = {6}^{3} . \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) = 216. \\ \\ = >{x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 6 = 216. \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } + 18 = 216. \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = 216 - 18. \\ \\ \large \boxed{ \boxed{ \bf \therefore {x}^{3} + \frac{1}{ {x}^{3} } = 198.}}$


✔✔ Hence, it is solved ✅✅.

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THANKS


#BeBrainly.

Answer 2

Given : x = 3 - √8.

$= > \frac{1}{x} = \frac{1}{3 - \sqrt{8}}* \frac{3+\sqrt{8}} {3 +\sqrt{8}}$

$=> \frac{1}{x} =\frac{3+\sqrt{8}}{(3)^2-(\sqrt{8})^2}$

$=> \frac{1}{x} =\frac{3+\sqrt{8}}{9 -8}$

$=> \frac{1}{x} =3+\sqrt{8}$

Now,

⇒ x + 1/x = 3 - √8 + 3 + √8

              = 6.

Hence, x + 1/x = 6.

On cubing both sides, we get

⇒ (x + 1/x)^3 = (6)^3

⇒ x^3 + (1/x^3) + 3(x + 1/x) = 216

⇒ x^3 + (1/x^3) + 3(6) = 216

⇒ x^3 + (1/x^3) = 216 - 18

⇒ x^3 + 1/x^3 = 198.

Therefore:

$=>\boxed {x^3 + \frac{1}{x^3} = 198}}$

Hope it helps!