Task 2. Replace the asterisk (*) in the expression $\left(x^{4}-3\right)^{2}+\left(x^{3}+*\right)^{2}$ with a monomial so that after squaring and combining like terms, there are four terms.
# Solution.

Let's replace $*$ with $3 x:$

$\left(x^{4}-3\right)^{2}+\left(x^{3}+3 x\right)^{2}=x^{8}-6 x^{4}+9+x^{6}+6 x^{4}+9 x^{2}=x^{8}+x^{6}+9 x^{2}+9$

Another answer is possible, for example, $\sqrt{6} x^{2}$

Analysis of the solution: Let's analyze the situation assuming that the coefficient of a monomial can be negative (a term can be a monomial with a negative coefficient).

After expanding the first parenthesis, three terms are formed: $x^{8} ;-6 x^{4} ; 9$. Let's see how the sum can change after expanding the second parenthesis:

1) If the sum in the second parenthesis is 0, there will only be three terms.
2) If the sum in the second parenthesis is a non-zero monomial (i.e., * is replaced by a monomial of the form $a x^{3}$, where $a \neq-1$), then a monomial of degree 6 will be added to the total sum, and since there are no such terms among the existing ones, there will be four terms: $x^{8},-6 x^{4}, 9,(a+1)^{2} x^{6}$
3) If the degree of the monomial is not equal to 2, after squaring the second parenthesis, three more terms will appear, making a total of six terms. To have only four terms after combining like terms, either one of the three new terms must cancel out with one of the three existing ones, or the degrees of two terms from the first three must match two terms from the second. Clearly, nothing can cancel out with 9. If the term $x^{8}$ cancels out, then $*=-x^{5} / 2$. If the term $-6 x^{4}$ cancels out, then $*=3 x$ or $*=\sqrt{6} x^{2}$.

Let's consider the case where two pairs of degrees match. The degrees of the terms in the first three are 0, 4, 8. In the second three, the degrees are $3, \kappa+3,2 \kappa$ (where $\kappa$ is the degree of *). Clearly, $2 \kappa$ and $\kappa+3$ cannot simultaneously equal any two of the numbers $0,4,8$.

Thus, * can be replaced by monomials $a x^{3}, a \neq-1, -x^{5} / 2, 3 x, \sqrt{6} x^{2}$, and in all these cases, after combining like terms, the final sum will consist of four terms. Note that this will be a sum with positive coefficients only in the case of replacing * with $3 x$ or $\sqrt{6} x^{2}$.

## Grading criteria:

- For a correct solution, it is sufficient to provide one correct answer and verify that it fits. Correct solution - 7 points.
- Correctly found monomial. Correct solution with one error (typo when squaring - 4 points.
- Found monomial, but not explained why it is a solution - 2 points.