Question

30. If x=2-√3, find the value of (i) x^3+1/x^3 (ii) x^2+1/x^2​

Answer 1

Answer:

Hey... here is your answer in the image.

Hope it helps! ! !

Answer 2

$\mathfrak{\large{\underline{\underline{Answer :-}}}}$

i) x³ + 1/x³ = 52

ii) x² + 1/x² = 14

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :

$x = 2 - \sqrt{3}$

To find :

$1) {x}^{3} + \dfrac{1}{ {x}^{3} } \\ \\ 2) {x}^{2} + \dfrac{1}{ {x}^{2} }$

Solution :

First find the value of 1/x

First find the value of 1/x To find the value of 1/x you need to rationalize the denominator

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

The rationalising factor of 2 - √3 is 2 + √3. So, multiply the numerator and denominator with rationalising factor.

$= \dfrac{1}{2 - \sqrt{3} } \times \dfrac{2 + \sqrt{3} }{2 + \sqrt{3} }$

$= \dfrac{2 + \sqrt{3} }{ {2}^{2} - {( \sqrt{3})}^{2} }$

Since (x + y)(x - y) = x² - y²

$= \dfrac{2 + \sqrt{3} }{4 - 3}$

$= \dfrac{2 + \sqrt{3} }{1}$

$= 2 + \sqrt{3}$

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

Now find the value of 1/x³

Now find the value of 1/x³ 1/x³ is nothing but (1/x)³

${(\dfrac{1}{x})}^{3} = {(2 + \sqrt{3})}^{3}$

$= {2}^{3} + 3( {2}^{2})( \sqrt{3}) + 3(2){( \sqrt{3})}^{2} + {( \sqrt{3})}^{3}$

Since (x + y)³ = x³ + 3x²y + 3xy² + y³

$= 8 + 3(4)( \sqrt{3}) + 6(3) + {( \sqrt{3})}^{2}( \sqrt{3} )$

$= 8 + 12 \sqrt{3} + 18 + 3 \sqrt{3}$

$= 26 + 15 \sqrt{3}$

${( \dfrac{1}{x})}^{3} = 26 + 15 \sqrt{3}$

Find the value of x³

${x}^{3} = {(2 - \sqrt{3})}^{3}$

Since (x - y)³ = x³ - 3x²y + 3xy² - y³

$= {2}^{3} - 3( {2}^{2})( \sqrt{3}) + 3(2){( \sqrt{3})}^{2} - {( \sqrt{3})}^{3}$

$= 8 - 3 \sqrt{3} (4) + 6(3) - { (\sqrt{3})}^{2}( \sqrt{3})$

$= 8 - 12 \sqrt{3} + 18 - 3 \sqrt{3}$

$= 26 - 15 \sqrt{3}$

${x}^{3} = 26 - 15 \sqrt{3}$

Now you can find the value of x³ + 1/x³

${x}^{3} - \dfrac{1}{ {x}^{3} } = 26 - 15 \sqrt{3} + (26 +15 \sqrt{3})$

$= 26 - 15 \sqrt{3} + 26 + 15 \sqrt{3}$

$= 52$

${x}^{3} + \dfrac{1}{ {x}^{3}} = 52$

Now find the value of 1/x²

Now find the value of 1/x²1/x² is nothing but (1/x)²

${(\dfrac{1}{x})}^{2} = {(2 + \sqrt{3}) }^{2}$

$= {2}^{2} + 2(2)( \sqrt{3}) + {( \sqrt{3})}^{2}$

Since (x + y)² = x² + 2xy + y²

$= 4 + 4 \sqrt{3} + 3$

$= 7 + 4 \sqrt{3}$

$\dfrac{1}{ {x}^{2} } = 7 + 4 \sqrt{3}$

Find the value of x²

${x}^{2} = {(2 - \sqrt{3})}^{2}$

$= {2}^{2} - 2(2)( \sqrt{3}) + {( \sqrt{3})}^{2}$

Since (x - y)² = x² - 2xy + y²

$= 4 - 4 \sqrt{3} + 3$

$=7 - 4 \sqrt{3}$

${x}^{2} = 7 - 4 \sqrt{3}$

Now you can find the value of x² + 1/x²

${x}^{2} - \dfrac{1}{ {x}^{2} } = 7 - 4 \sqrt{3} + (7 + 4 \sqrt{3})$

$= 7 - 4 \sqrt{3} + 7 + 4 \sqrt{3}$

$= 7 + 7$

$= 14$

${x}^{2} + \dfrac{1}{ {x}^{2} } = 14$