Question

Pls solve this question ​

Answer 1

Question :

$\sf If \: x + \frac{1}{x} = 2 , \: prove \: that \: {x}^{2} + \frac{1}{ {x}^{2} } = {x}^{3} + \frac{1}{ {x}^{3} } = {x}^{4} + \frac{1}{ {x}^{4} }$

Formula Used :

For x² + 1/x² we will use :

$\sf \looparrowright {x}^{2} + \frac{1}{ {x}^{2} } = {(x + \frac{1}{x}) }^{2} - 2$

For x³ + 1/x³ we will use :

$\sf \looparrowright {x}^{3} + \frac{1}{ {x}^{3} } = {(x + \frac{1}{x}) }^{3} - 3(x + \frac{1}{x} )$

For x⁴ + 1/x⁴ we will use :

$\sf \looparrowright {{x}^{4} + \frac{1}{ {x}^{4}}} = {( { {x}^{2} } + \frac{1}{ {x}^{2} } )^{2} } - 2$

GivEn Data :

x + 1/x = 2

Solution :

By applying the first formula we get

$\sf \leadsto {x}^{2} + \frac{1}{ {x}^{2} } = {(x + \frac{1}{x}) }^{2} - 2$

By putting 2 we get

$\sf \leadsto {x}^{2} + \frac{1}{ {x}^{2} } = {(2) }^{2} - 2$

$\sf \leadsto {x}^{2} + \frac{1}{ {x}^{2} } = 4 - 2$

$\sf \leadsto {x}^{2} + \frac{1}{ {x}^{2} } = 2$

Now, to apply the second formula :

$\sf \leadsto {x}^{3} + \frac{1}{ {x}^{3} } = {(x + \frac{1}{x}) }^{3} - 3(x + \frac{1}{x} )$

$\sf \leadsto {x}^{3} + \frac{1}{ {x}^{3} } = {(2) }^{3} - 3(2)$

$\sf \leadsto {x}^{3} + \frac{1}{ {x}^{3} } = 8 - 6$

$\sf \leadsto {x}^{3} + \frac{1}{ {x}^{3} } = 2$

Now to find the last one using the third formula :

$\sf \leadsto {{x}^{4} + \frac{1}{ {x}^{4}}} = {( { {x}^{2} } + \frac{1}{ {x}^{2} } )^{2} } - 2$

By putting the value we got in x² + 1/x² we get

$\sf \leadsto {{x}^{4} + \frac{1}{ {x}^{4}}} = {( 2 )^{2} } - 2$

$\sf \leadsto {x}^{4} + \frac{1}{ {x}^{4} } = 4 - 2$

$\sf \leadsto {{x}^{4} + \frac{1}{ {x}^{4}}} = 2$

So we got the values as

$\leadsto \boxed{ \sf {x}^{2} + \frac{1}{ {x}^{2} } = 2 }$

$\leadsto \boxed{ \sf {x}^{3} + \frac{1}{ {x}^{3} } = 2}$

$\leadsto \boxed{\sf {{x}^{4} + \frac{1}{ {x}^{4}}} = 2}$

Hence, we got

$\sf \rarr \red{{x}^{2} + \frac{1}{ {x}^{2}} } = \blue{{x}^{3} + \frac{1}{ {x}^{3} }} = \green{ {{x}^{4} + \frac{1}{ {x}^{4}}} }= \boxed{ \sf 2}$

$\mathcal{\underbrace{ \red{HENCE, \: PROVED}} }$

KNOW MORE :

• (a + b)² = a² + b² + 2ab

• (a - b)² = a² + b² - 2ab

• a² + b² = (a + b)² - 2ab

• a² + b² = (a - b)² + 2ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)² - (a - b)² = 4ab

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a + b)³ = a³ - b³ - 3ab(a + b)

• a³ - b³ = (a + b)³ + 3ab(a + b)

Regards

#BeBrainly