Finding cubic with golden ratio as root

If the coefficients are rational and $\dfrac{1+\sqrt 5} 2$ is a root, then $\dfrac{1-\sqrt 5} 2$ is a root.

To see this, suppose you substitute $\dfrac{1+\sqrt 5} 2$ for $x$ and get $0$. What would then happen if you substitute $\dfrac{1-\sqrt 5} 2$ for $x$? When you expand $x^2$ and $x^3$, then wherever $\sqrt 5$ appears, $-\sqrt 5$ would appear, and vice-versa. You won't get $0$ unless the coefficient of $\sqrt 5$ in the total ends up being $0$. If you interchange $\pm\sqrt 5$ then instead of $0$ you get $-0$, but $-0$ is $0$.

The fact that $\sqrt 5$ is irrational is essential here. Suppose $\sqrt 5$ were the rational number $38/17$. Then $17x-38$ would be a polynomial with integer coefficients having $\sqrt 5$ as a root. The argument in the paragraph above assumes the radical cannot vanish like that.

This is very much like the proof that if the coefficient are real and $a+bi$ is a root, where $a$ and $b$ are real, then $a-bi$ is also a root.

No such cubic exists.

The golden ratio is a root of the quadratic $q(x)=x^2-x-1$.

If the golden ratio is a root of a cubic $p(x)$, then it must be the root of the remainder of $p(x)$ when divided by $q(x)$. That remainder has integer coefficients and is either zero or has degree $1$. But the golden ratio is not a root of polynomial of degree $1$ with integer coefficients because it is irrational. Thus the remainder is zero and the cubic must have at least two real roots: those of $q(x)$.

Note that the third condition in the question is not used in this argument.