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Discuss Ian's answer to: If (x^2+1/x^2)=7 then (x^5+1/x^5)

Let x^5 + 1/(x^5) = T, and

x + 1/x = t, so that

t^2 = x^2 + 1/(x^2) + 2 = 7 + 2 = 9                            (a)

Note that t^2 is single valued, but t = +/- 3

t^3 = x^3 + 3x + 3/x + 1/(x^3) = [x^3+ 1/(x^3)] + 3t    (b)

t^5 = x^5 + 5x^3 + 10x + 10/x + 5/(x^3)+ 1/(x^5)        (c)

t^5 = T + 5(t^3 - 3t) +10t      by substituting (b) into (c)

T = t^5 - 5t^3 +5t = t[(t^2)^2 - 5t^2 + 5]

T = (+/- 3)(81 - 45 + 5) = +/- 123

 

(Or use trig expansions z^n + 1/(z^n) = 2 cos (n theta)

with z = e^i theta )

Hope this was helpful,

Regards - Ian

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