Given $x \in \mathbb R$ such that $8x^3 - 4x^2 - 4x + 1 = 0$. Find all rationals $p, q, r$ such that $px + q(2x^2 - 1) + r(4x^3 - 3x) = -4$. 
Given $x \in \mathbb R$ such that $8x^3 - 4x^2 - 4x + 1 = 0$. Find all rationals $p, q, r$ such that $$\large px + q(2x^2 - 1) + r(4x^3 - 3x) = -4$$

We have that $$8x^3 - 4x^2 - 4x + 1 = 0 \iff (-1) + 2x - 2 \cdot (2x^2 - 1) + 2 \cdot (4x^3 - 3x) = 0$$ and $$px + q(2x^2 - 1) + r(4x^3 - 3x) = -4 \iff (-4) \cdot (-1) + px + q(2x^2 - 1) + r(4x^3 - 3x) = 0$$
This implies that $(p + 8)x + (q - 8)(2x^2 - 1) + (r + 8)(4x^3 - 3x) = 0$.
One solution is that $p + 8 = q - 8 = r + 8 = 0 \implies (p, q, r) = (-8, 8, -8)$
I suspect to there be other solutions of $(p, q, r)$, for example, $$x = \frac{1}{6} \cdot \left[1 - \sqrt[3]{\frac{7}{2}(3\sqrt3i + 1)} + \sqrt[3]{\frac{7}{2}(3\sqrt3i - 1)}\right]$$, according to WolframAlpha, is a solution to the equation $8x^3 - 4x^2 - 4x + 1 = 0$.
Right... So WolframAlpha is untrustworthy, according to Michael Rozenberg, the solutions to $8x^3 - 4x^2 - 4x + 1 = 0$ are $\cos\dfrac{\pi}{7}, \cos\dfrac{3\pi}{7}, \cos\dfrac{5\pi}{7}$, which is correct when plugged into the original equation.

$8x^3 - 4x^2 - 4x + 1 = 0$ and  $4rx^3+2qx^2+(p-3r)x+(4-q)=0$.
Case 1
If the original cubic is irreducible over the rationals then the coefficients of the various powers of $x$ in the two cubics must be in proportion. (Otherwise, one or more roots would satisfy a rational polynomial of degree less than $3$ obtained as a linear combination of the two cubics.) Then 
$$4r=8t, 2q=-4t,p-3=-4t,4-q=t.$$
This gives the solution $p=-8,q=8,r=-8.$
Case 2
Otherwise the cubic has a rational root $\frac{a}{b}$ with $(a,b)=1$ and so 
$$8a^3-4a^2b-4ab^2+b^3=0$$ Then $a$ divides $b^3$ and so $a=1$. Then $b$ would have to be $\pm2$ but these do not satisfy the equation and so there are no further solutions. 
