1. If `x^2+1/x^2 = 2`, then find `x-1/x`

Solution:
Here `x^2+1/x^2=2`


Now, We know that
`(x-1/x)^2=(x^2+1/x^2)-2`

`:.(x-1/x)^2=2-2`

`:.(x-1/x)^2=0`

`:.x-1/x=0`

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2. If `x^3+1/x^3 = 2`, then find `x+1/x`

Solution:
Here `x^3+1/x^3=2`


Now, We know that
`(x+1/x)^3=(x^3+1/x^3)+3(x+1/x)`

`:.(x+1/x)^3-3(x+1/x)=(x^3+1/x^3)`

`:.(x+1/x)^3-3(x+1/x)=2`

`:.y^3-3y -2=0`, where `y=(x+1/x)`

clearly, y=2 satisfies `y^3-3y -2=0`

`:.x+1/x=2`

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3. If `x^3-1/x^3 = 2`, then find `x-1/x`

Solution:
Here `x^3-1/x^3=2`


Now, We know that
`(x-1/x)^3=(x^3-1/x^3)-3(x-1/x)`

`:.(x-1/x)^3+3(x-1/x)=(x^3-1/x^3)`

`:.(x-1/x)^3+3(x-1/x)=2`

`:.y^3+3y -2=0`, where `y=(x-1/x)`

clearly, y=2 satisfies `y^3+3y -2=0`

`:.x-1/x=2`

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4. If `x^4+1/x^4 = 2`, then find `x-1/x`

Solution:
Here `x^4+1/x^4=2`


Now, We know that
`(x^2+1/x^2)^2=(x^4+1/x^4)+2`

`:.(x^2+1/x^2)^2=2+2`

`:.(x^2+1/x^2)^2=4`

`:.x^2+1/x^2=2`


Now, We know that
`(x-1/x)^2=(x^2+1/x^2)-2`

`:.(x-1/x)^2=2-2`

`:.(x-1/x)^2=0`

`:.x-1/x=0`

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This material is intended as a summary. Use your textbook for detail explanation.
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