Find another polynomial with the same roots as a polynomial with radical coefficients

Question

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 . $$\sqrt{7}(x^3 - 1) + 1 = \sqrt{5}(x^3 - x)$$ $$7(x^3 - 1)^2 + 2\sqrt{7}(x^3 - 1) + 1 = 5(x^3 - x)^2$$ $$2\sqrt{7}(x^3 - 1) = 5(x^3 - x)^2 - 7(x^3 - 1)^2 -1$$ $$28(x^3 - 1)^2 = (5(x^3 - x)^2 - 7(x^3 - 1)^2 -1)^2$$