You can certainly do it using Vieta's formulas, but it will be laborious. I don't see any simpler way, though.

Let be the roots of . The Vieta formulas tell you that , and . You want to find the equation whose roots are and .

The coefficients of that equation are again given by the Vieta equations. For example, and . Those two are easy, but the others are harder. For example, . To evaluate that, notice that , and it follows that . I haven't even tried to do or , but it should be possible, using the standard relations among symmetric polynomials.