The idea is basically:
Any monic polynomial can be factored as $f(x) = \prod (x - a_i)$, where $a_{1,\dots,n}$ are the roots of the polynomial.
Now if we expand such a product:
$(x - a_1)(x - a_2) = x^2 - (a_1 + a_2)x + a_1a_2$
$(x - a_1)(x - a_2)(x - a_3) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_1a_3 + a_2a_3)x - a_1a_2a_3$
And so on. The pattern should be clear.
This means that finding the roots of a polynomial is in fact equivalent to solving systems like the following:
For a quadratic polynomial $x^2 - px + q$, find $a_1,a_2$, such that
$p = a_1 + a_2$
$q = a_1a_2$
For a cubic polynomial $x^3 - px^2 + qx - r$, find $a_1,a_2,a_3$, such that
$p = a_1 + a_2 + a_3$
$q = a_1a_2 + a_1a_3 + a_2a_3$
$r = a_1 a_2 a_3$
And similarly for higher degree polynomials.
Not surprisingly, the amount of "unfolding" that needs to be done to solve the quadratic system is much less than the amount of "unfolding" needed for the cubic system.
The reason why polynomials of degree 5 or higher are not solvable by radicals, can be thought of as: The structure (symmetries) of the system for such a polynomial just doesn't match any of the structures that can be obtained by combining the structures of the elementary operations (adding subtracting, multiplication, division, and taking roots).
Note that the given equation
$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$
is purposefully structured to allow convenient factorization as follows
$$3x^3=x^4-(\sqrt{18}x+\sqrt{32})^2 -4x^2$$
$$x^4-3x^3-4x^2-(\sqrt{18}x+\sqrt{32})^2 =0$$
$$x^4-(3x+4)x^2-2(3x+4)^2 =0$$
$$(x^2+(3x+4))(x^2-2(3x+4))=0$$
Best Answer
Solving $$x^4 + 4 a x^3 + 6 x^2 + 4 a x + 1 = 0$$ Divide all terms by $x^2$ $$x^2+4ax+6+\frac{4a}{x}+\frac{1}{x^2}=0$$ Osserve that we can write the previous equation as $$\left(x^2+\frac{1}{x^2}+2\right)+4a\left(x+\frac{1}{x}\right)+4=0$$ That is $$\left(x+\frac{1}{x}\right)^2+4a\left(x+\frac{1}{x}\right)+4=0$$ Set $w=x+\frac{1}{x}$. The equation becomes $$w^2+4aw+4=0$$ Solutions are $$w_1=-2a-2 \sqrt{a^2-1};\;w_2=-2a+2 \sqrt{a^2-1}$$ Bringing beack the $x$ we have the equations $$x+\frac{1}{x}=-2a-2 \sqrt{a^2-1};\;x+\frac{1}{x}=-2a+2 \sqrt{a^2-1}$$ Reordering etc $$x^2 + \left(2a+2 \sqrt{a^2-1}\right)x+1=0;\;x^2 + \left(2a-2 \sqrt{a^2-1}\right)x+1=0$$ Whose solutions are $$x_1=-a-\sqrt{a^2-1}-\sqrt{2 a^2-2+2 a\sqrt{a^2-1} };\;\\x_2=-a-\sqrt{a^2-1}+\sqrt{2 a^2-2+2 a\sqrt{a^2-1}};\;\\x_3=-a+\sqrt{a^2-1}-\sqrt{2 a^2-2-2 a\sqrt{a^2-1} };\;\\x_4=-a+\sqrt{a^2-1}+\sqrt{2 a^2-2-2 a\sqrt{a^2-1} }$$