Problem 5. The number $a$ is such that the equation $t^{3}+4 t^{2}+4 t+a=0$ has three real roots $x<y<z$. Determine the numerical values that the expression

$$
A=x^{3}-4 y^{2}-4 z^{2}-4 y-4 z+32
$$

can take.
Answer: $\left(14 \frac{22}{27} ; 16\right)$.

Solution. Let $P(t)=t^{3}+4 t^{2}+4 t+a=(t-x)(t-y)(t-z)$. Then $x+y+z=-4, xy+$ $yz+xz=4, xyz=-a$ (this follows from Vieta's theorem or by directly comparing the left and right sides). Next, $A=x^{3}-4 y^{2}-4 z^{2}-4 y-4 z+32=\left(-4 x^{2}-4 x-a\right)-4 y^{2}-4 z^{2}-4 y-4 z+32=$ $-4\left(x^{2}+y^{2}+z^{2}\right)-4(x+y+z)-a+32=-4(x+y+z)^{2}+8(xy+yz+zx)-4(x+y+z)-a+32=$ $-4 \cdot(-4)^{2}+8 \cdot 4-4 \cdot(-4)-a+32=16-a$. It remains to determine for which values of $a$ the given cubic equation has three distinct roots. Write the equation in the form $a=-t^{3}-4 t^{2}-4 t=$ $-t(t+2)^{2}$ and find the extrema of the function on the right side (for example, using the derivative, although it can be done without it). It turns out that this function has a local minimum of 0 at the point $t=-2$ and a local maximum of $\frac{32}{27}$ at the point $t=-\frac{2}{3}$. This means that the equation will have 3 distinct roots for $a \in\left(0 ; \frac{32}{27}\right)$. Therefore, $A=16-a \in\left(16-\frac{32}{27} ; 16\right)=\left(14 \frac{22}{27} ; 16\right)=\left(\frac{400}{27} ; 16\right)$.