Question

if xnot=0 x+1/x=2 show

x2+1/x2=x3+1/x3=x4+1/x4

Answer 1

Answer :

Given x $\ne$ 0 And

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

Show

x² + $\frac{1}{x^2}$ = x³ + $\frac{1}{x^3}$ = x⁴ + $\frac{1}{x^4}$


Squaring equation 1 , we get

( x + ​$\frac{1}{x}$ )² = 2²

$\Rightarrow$x² + $\frac{1}{x^2}$ + 2 ( x ) ( $\frac{1}{x}$ ) = 4 [ As we know ( a + b )² = a² + b² + 2ab ]

$\Rightarrow$x² + $\frac{1}{x^2}$ + 2 = 4

$\Rightarrow$x² + $\frac{1}{x^2}$ = 4 - 2

$\Rightarrow$x² + $\frac{1}{x^2}$ = 2 ---------------------- ( 2 )

Now taking Cube of equation 1 , we get

( x + ​$\frac{1}{x}$ )³ = 2³

$\Rightarrow$x³ + $\frac{1}{x^3}$ + 3 ( x ) ( $\frac{1}{x}$ ) ( x + $\frac{1}{x}$ ) = 8 [ As we know ( a + b )³ = a³ + b³ + 3ab( a + b ) ]

$\Rightarrow$x³ + $\frac{1}{x^3}$ + 3 ( 2 ) = 8 ​

$\Rightarrow$x³ + $\frac{1}{x^3}$ + 6 = 8

$\Rightarrow$x³ + $\frac{1}{x^3}$ = 8 - 6

$\Rightarrow$x³ + $\frac{1}{x^3}$ = 2 ---------------------- ( 3 )

Now take power 4 of equation 1 , and get

( x + ​$\frac{1}{x}$ )⁴ = 2⁴

$\Rightarrow$x⁴ + $\frac{1}{x^4}$ + 4( x ) ( $\frac{1}{x}$ ) ( x + $\frac{1}{x}$ ) + 6( x² ) ($\frac{1}{x^2}$ ) = 16 [ As we know ( a + b )⁴ = a⁴ + b⁴ + 4ab( a + b ) + 6a² b² ]

$\Rightarrow$x⁴ + $\frac{1}{x^4}$ + 4 ( 2 ) + 6= 16 ​

$\Rightarrow$x⁴ + $\frac{1}{x^4}$ + 8 + 6= 16 ​​

$\Rightarrow$x⁴ + $\frac{1}{x^4}$ + 14 = 16 ​​


$\Rightarrow$x⁴ + $\frac{1}{x^4}$ = 16 ​​​- 14

$\Rightarrow$x⁴ + $\frac{1}{x^4}$ = 2 ---------------------- ( 4 )

Hence
So from equation 2 and equation 3 and equation 4 , we get

x² + $\frac{1}{x^2}$ = x³ + $\frac{1}{x^3}$ = x⁴ + $\frac{1}{x^4}$ ( Hence proved )