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I knew trick to solve these problems, but I forgot.

1. x^8 + \frac{1}{x^8} = 2207. Find value of x + \frac{1}{x} .

2. x + \frac{1}{x} = 2. Find value of x^9 + \frac{1}{x^9} .

I tried to find similar problems on forum, but could not.

i think you can write it as X^8 + X^(-8) = 2207 and then just take 2207^-8 to get it to the desired state? The reverse for the 2nd one where you just do 2^9 to get from x^1 + x^(-1) to x^9 + x^-9. it is a bit rusty on me too.... _________________

I was able to solve second Q. We just simply take cube & middle terms are replaceable.

In first Q, 2207 is a prime number. So, we can't take it's square root or anything like that. I am still not able to find solution for that. I just know the answer for that problem - 3. _________________

Re: Algebra Basics. [#permalink]
21 Dec 2009, 15:30

I think I can help

If we suppose that x+\frac{1}{x}=b

then we can find: (x+\frac{1}{x})^2=b^2 x^2+2+\frac{1}{x^2}=b^2 x^2+\frac{1}{x^2}=b^2-2 (x^2+\frac{1}{x^2})^2=(b^2-2)^2 x^4+2+\frac{1}{x^4}=(b^2-2)^2 x^4+\frac{1}{x^4}=(b^2-2)^2-2 (x^4+\frac{1}{x^4})^2=((b^2-2)^2-2)^2 x^8+2+\frac{1}{x^8}=((b^2-2)^2-2)^2 x^8+\frac{1}{x^8}=((b^2-2)^2-2)^2-2

and now to find x+\frac{1}{x} we will solve ((b^2-2)^2-2)^2-2=2207

((b^2-2)^2-2)^2=2209=47^2 (b^2-2)^2-2=47 (b^2-2)^2=49 b^2-2=7 b^2=9 b=3 or b=-3

and it means that x+\frac{1}{x}=3 or x+\frac{1}{x}=-3.

To solve second problem it is enough to exponentiate the equation twice:

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