Question

Please solve q 10 please

Answer 1

Answer:

hey mate here is ur answer hope u find it helpful

Answer 2

☞ Show that x³ + $\dfrac{1}{ {x}^{3} }$ = 52

x = 2 + √3 ______[GIVEN]

Solution:

=> x = 2 + √3

=> $\dfrac{1}{x}$ = $\dfrac{1}{2 \: + \: \sqrt{3} }$

• Rationalize

=> $\dfrac{1}{x}$ = $\dfrac{1}{2 \: + \: \sqrt{3} }$ × $\dfrac{2 \: - \: \sqrt{3} }{2 \: - \: \sqrt{3} }$

=> $\dfrac{1}{x}$ = $\dfrac{2 \: - \: \sqrt{3} }{( {2}^{2}) \: - \: { (\sqrt{3} )}^{2} }$

=> $\dfrac{1}{x}$ = $\dfrac{2 \: - \: \sqrt{3} }{4 \: - \: 3}$

=> $\dfrac{1}{x}$ = $\dfrac{2 \: - \: \sqrt{3} }{1}$ _______(eq 1)

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To show x³ + $\dfrac{1}{ {x}^{3} }$ = 52.

Let us find before x + $\dfrac{1}{x}$

We know that x = 2 + √3 [GIVEN]

and $\dfrac{1}{x}$ = $\dfrac{2 \: - \: \sqrt{3} }{1}$ [FROM (eq 1)]

=> x + $\dfrac{1}{x}$ = 2 + √3 + 2 - √3

=> x + $\dfrac{1}{x}$ = 4 ________(eq 2)

• Take cube on both sides.

=> $(x \: + \: { \dfrac{1}{x} )}^{3}$ = (4)³

=> (x)³ + $\dfrac{1}{ {x}^{3} }$ + 3(x + $\dfrac{1}{x}$) = 64

=> (x)³ + $\dfrac{1}{ {x}^{3} }$ + 3 × 4 = 64 [From (eq2)]

=> (x)³ + $\dfrac{1}{ {x}^{3} }$ + 12 = 64

=> (x)³ + $\dfrac{1}{ {x}^{3} }$ = 64 - 12

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(x)³ + $\dfrac{1}{ {x}^{3} }$ = 52

____________[HENCE PROVED]