<s>[INST] integer $a$ for which $x^3-3x+a = 0$ has $3$ distinct real roots in $x\in (0,1)$ If the equation $x^3-3x+a=0$ has $3$ distinct roots between $0$ and $1$ then the integer value of $a$ is
Can we solve it without Calculus.
$\bf{My\; Try}::$ Let $x=2\cos \phi$, where $-1\leq \cos \phi\leq 1\Rightarrow 0\leq \phi \leq \pi$(Principle value, Inverse exists)
$\displaystyle 8\cos^3 \phi-6\cos \phi+a = 0\Rightarrow 4\cos^3 \phi-3\cos \phi = -\frac{a}{2}\Rightarrow \cos 3\phi = -\frac{a}{2}$
Now $-1\leq \cos 3\phi\leq 1$, So $\displaystyle -1 \leq -\frac{a}{2}\leq 1\Rightarrow -2\leq a\leq 2$
So we get $a = \left\{-2,-1,0,1,2\right\}$
But answer is no integer value of $a$
Help me
Thanks
 [/INST] If we factor the polynomial as $(x - b)(x - c)(x - d)$, we have
$$a = -bcd$$
But since $0 < b, c, d < 1$, we have that $|a| < 1$; the only integer satisfying this is zero, which certainly doesn't work.
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