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If $\sqrt{7}(x^3 - 1) - \sqrt{5}(x^3 - x) + 1 = 0$ is an expression, how would you get a similar expression with the same roots as the one above but without irrational coefficients? I tried bringing it over like $\sqrt{7}(x^3 - 1) + 1 = \sqrt{5}(x^3 - x)$ and squaring it and doing other step but I keep on getting stuck. Any advice? Edit: This is what I have so far, I continued simplifying, but on wolfram, it was expressed as having different roots so I just stopped there . $$\sqrt{7}(x^3 - 1) + 1 = \sqrt{5}(x^3 - x)$$ $$7(x^3 - 1)^2 + 2\sqrt{7}(x^3 - 1) + 1 = 5(x^3 - x)^2$$ $$2\sqrt{7}(x^3 - 1) = 5(x^3 - x)^2 - 7(x^3 - 1)^2 -1$$ $$28(x^3 - 1)^2 = (5(x^3 - x)^2 - 7(x^3 - 1)^2 -1)^2$$

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  • $\begingroup$ Starting with $$\sqrt{7}(x^3-1)+1=\sqrt{5}(x^3-x)$$Square both sides to get$$2\sqrt{7}(x^3-1)=-2x^6-10x^4+14x^3+5x^2-8$$Then square again. The resulting polynomial will be a degree of $12$. Is this really what you want? $\endgroup$ – Crescendo Mar 3 '17 at 1:56
  • $\begingroup$ yeahh thats what i tried doing. i eventually got $(5(x^3-x)^2-7(x^3-1)^2-1)^2 - 25(7)(x^3-1)^2$ but when I plugged it into wolfram, there appeared to be distinct roots. $\endgroup$ – kuhe Mar 3 '17 at 1:59
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    $\begingroup$ @kuhe: If you want rational coefficients, you will have to accept extraneous solutions (i.e., roots which don't satisfy the original equation). That's just the nature of the game. $\endgroup$ – quasi Mar 3 '17 at 2:32
  • $\begingroup$ As @quasi says, it’s the nature of the game, and the reason for this is that algebraically, $\sqrt7$ and $-\sqrt7$ behave identically. Similarly, $\sqrt5$ and $-\sqrt5$. $\endgroup$ – Lubin Mar 3 '17 at 4:09
  • $\begingroup$ At least this OP beat their classmates to it :-) $\endgroup$ – Jyrki Lahtonen Mar 3 '17 at 8:03

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