2021/060) Find a polynomial of degree 3 with real coefficients such that each of its roots is equal to the square of one root of the polynomial $P(x)=x^3+9x^2+9x+9$.

Because the square of the roots shall be roots of the required polynomial so we must have $\sqrt x$ as roots of P(X)

So $(\sqrt x)^3 + 9 x + 9 (\sqrt x) + 9 = 0$

Or $(\sqrt x)(x + 9) = - 9(x+1)$

Or squaring $x(x+9)^2 = 81(x+1)^2$

Or $x(x^2 + 18 x + 81) = 81 x^2 + 162x + 81$

Or $x^3 - 63x^2 - 81 x - 81 = 0$

This is the required equation