Which of the following statements are true about the given rational equation? Check all of the boxes that apply. \( \begin{array}{l}\frac{4}{x+6}+\frac{1}{x^{2}}=\frac{x+10}{x^{3}+6 x^{2}} \\ x=1 \text { is a solution. } \\ x=0 \text { is a solution. } \\ x=-1 \text { is a solution. } \\ x=-6 \text { is a solution. }\end{array} \).

First, let's clarify that to determine the truth of these statements, we need to identify valid solutions for the given rational equation. We can start by simplifying the equation and substituting the proposed values (1, 0, -1, -6) to see whether they satisfy the equation. Evaluating \( x = 1 \): \[ \frac{4}{1 + 6} + \frac{1}{1^2} = \frac{4}{7} + 1 \neq \frac{1 + 10}{1^3 + 6 \cdot 1^2} \text{ (not a solution)} \] Evaluating \( x = 0 \): \[ \frac{4}{0 + 6} + \frac{1}{0^2} \text{ is undefined (not a solution)} \] Evaluating \( x = -1 \): \[ \frac{4}{-1 + 6} + \frac{1}{(-1)^2} = \frac{4}{5} + 1 \neq \frac{-1 + 10}{(-1)^3 + 6(-1)^2} \text{ (not a solution)} \] Evaluating \( x = -6 \): \[ \frac{4}{-6 + 6} + \frac{1}{(-6)^2} \text{ is undefined (not a solution)} \] Now, we conclude that all proposed values are not solutions. In summary: - The first statement \( x=1 \text{ is a solution} \) is false. - The second statement \( x=0 \text{ is a solution} \) is false. - The third statement \( x=-1 \text{ is a solution} \) is false. - The fourth statement \( x=-6 \text{ is a solution} \) is false. None of the boxes should be checked; all statements are false.