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प्रश्न
If X ≠ 0 and X + `1/"X"` = 2 ; then show that :
`x^2 + 1/x^2 = x^3 + 1/x^3 = x^4 + 1/x^4`
उत्तर
`( x + 1/x )^2 = x^2 + 1/x^2 + 2`
⇒ `x^2 + 1/x^2 = ( x + 1/x )^2 - 2`
⇒ `x^2 + 1/x^2 = (2)^2 - 2 [ ∵ x + 1/x = 2 ]`
⇒ `x^2 + 1/x^2 = 2` .....(1)
`( x + 1/x )^3 = x^3 + 1/x^3 + 3( x + 1/x)`
⇒ `x^3 + 1/x^3 = ( x + 1/x )^3 - 3( x + 1/x )`
⇒ `x^3 + 1/x^3 = (2)^3 - 3(2) [ ∵ x + 1/x = 2 ]`
⇒ `x^3 + 1/x^3 = 8 - 6`
⇒ `x^3 + 1/x^3 = 2` ...(2)
We know that
`x^4 + 1/x^4 = ( x^2 + 1/x^2 )^2 - 2`
= `(2)^2 - 2` [ from (1) ]
= 4 - 2
⇒ `x^4 + 1/x^4 = 2` ...(3)
Thus from equations (1), (2) and (3), we have
`x^2 + 1/x^2 = x^3 + 1/x^3 = x^4 + 1/x^4`
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