Hi there!

How does one factorize $\displaystyle x^3 + (x^-3) - 2$?

I tried factorizing it by $\displaystyle x^3 - 1 + (x^-3) -1$ and a few other methods

but couldn't really get close to the answer.

Thanks

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- May 13th 2013, 11:04 PMMoniMiniFactorization
Hi there!

How does one factorize $\displaystyle x^3 + (x^-3) - 2$?

I tried factorizing it by $\displaystyle x^3 - 1 + (x^-3) -1$ and a few other methods

but couldn't really get close to the answer.

Thanks - May 13th 2013, 11:42 PMibduttRe: Factorization
let x^3 = a

The equation becomes a + 1/a -2 = 0

a^2 - 2a + 1 = 0 That gives ( a -1 )^2 = 0 OR ( x^3 - 1 )^2 = 0

and we know the factors of x^3 - 1 - May 14th 2013, 07:22 AMSorobanRe: Factorization
Hello, MoniMini!

Another approach . . .

Quote:

$\displaystyle \text{Factor: }\: x^3 + x^{-3} - 2$

We have: .$\displaystyle x^3 - 2 + \frac{1}{x^3} \;=\;\frac{x^6 - 2x^3 + 1}{x^3} \;=\;\frac{(x^3-1)^2}{x^3}$

. . . . . . . $\displaystyle =\;\frac{[(x-1)(x^2+x+1)]^2}{x^3} \;=\; \frac{(x-1)^2(x^2+x+1)^2}{x^3}$