show that all the roots of a given equation are in $[-1,1]$ 

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*Suppose $c\in \mathbb{R}$ is such that one of the roots of the equation $x^3 - \dfrac{3}4 x+c=0$ belongs to $[-1,1]$.

*Show that all the roots belong to $[-1,1]$.

*The equation is equivalent to $p(x)=0$ where $p(x) = 4x^3 - 3x+4c.$


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*Observe that $4\cos^3\theta - 3\cos \theta = \cos(3\theta).$

*Let the root $r$ of the equation in $[-1,1]$ be equal to $\cos \theta$ for some $\theta \in [0,\pi]$.

*From the above observation, $c = -\dfrac{\cos \theta_2}4$ for some $\theta_2 \in [0,\pi]$.

*The equation has two more roots, and if one of them is complex, then because the polynomial $p(x)$ has real coefficients, both of the others must be complex. We know that if $r_2,r_3$ are the other roots, then $r r_2 r_3 = -c, r + r_2 + r_3 = 0, r r_2 + r r_3 + r_2r_3 = -3/4.$
I'm not sure how to derive a contradiction from here.

 A: Without calculus:
Let $r$ be a root of the equation in $[-1, 1]$. We have $c = -r^3 + \frac34 r$.
The equation becomes
$$x^3 - \frac34x - r^3 + \frac34 r = 0$$
or
$$\frac14(x-r)(4x^2 + 4rx + 4r^2 - 3) = 0$$
which has exactly three real roots
$$x_1 = r,\quad x_2 = -\frac12 r + \frac12\sqrt{3 - 3r^2}, \quad
x_3 = -\frac12 r - \frac12\sqrt{3-3r^2}.$$
It is easy to prove that $x_2, x_3 \in [-1, 1]$.
Indeed, first, we have $x_2 \ge - \frac12 r \ge -\frac12$
and $x_3 \le -\frac12 r \le \frac12$.
Second, we have
$$1 - x_2 = 1 + \frac12 r - \frac12\sqrt{3 - 3r^2}
\ge 0$$
since $1 + r/2 > 0$ and $1 + r + \frac14 r^2 - \frac{3 - 3r^2}{4} = (r + 1/2)^2 \ge 0$.
Third, we have
$$1 + x_3 = 1 - \frac12 r - \frac12\sqrt{3 - 3r^2} \ge 0$$
since $1 - r/2 > 0$ and $1 - r + \frac14r^2 - \frac{3-3r^2}{4} = (r - 1/2)^2 \ge 0$.
We are done.
A: $f(x)=x^3-3x/4\implies f'(x)=3(x^2-1/4)\implies f'(x)>0, \forall x \in (-\infty,-1/2) \cup (1/2,\infty).$ As $f(x)$ is a monotonically increasing when $|x|>1/2$, the equation $f(x)+c=0$ cannot have real roots for $|x|>1/2.$
