1. It is known that the equation $x^{4}-8 x^{3}+a x^{2}+b x+16=0$ has (taking into account multiplicity) four positive roots. Find $a-b$.
Solution: Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the roots of our equation (some of them may be the same). Therefore, the polynomial on the left side of the equation can be factored as:

$$
x^{4}-8 x^{3}+a x^{2}+b x+16=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)
$$

Expanding the brackets on the right side and equating the coefficients of like powers of $x$, we get:

$$
x_{1}+x_{2}+x_{3}+x_{4}=8, \quad x_{1} x_{2} x_{3} x_{4}=16
$$

It is known that the geometric mean of non-negative numbers does not exceed their arithmetic mean, but in our case, they are equal:

$$
\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}=\sqrt[4]{x_{1} x_{2} x_{3} x_{4}}=2
$$

Therefore, $x_{1}=x_{2}=x_{3}=x_{4}=2$, and

$$
x^{4}-8 x^{3}+a x^{2}+b x+16=(x-2)^{4}
$$

From this, $a=24, b=-32$.

Answer: 56