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

We start with the equation \[ \frac{4}{x+6}+\frac{1}{x^{2}}=\frac{x+10}{x^{3}+6x^{2}}. \] Notice that the denominator on the right factors as \[ x^{3}+6x^{2}=x^2(x+6). \] Thus, the equation can be rewritten as \[ \frac{4}{x+6}+\frac{1}{x^{2}}=\frac{x+10}{x^2(x+6)}. \] Since \(x^2(x+6)\) is the common denominator, we multiply both sides by it (remembering that \(x \ne 0\) and \(x \ne -6\) to avoid division by zero): \[ \left(\frac{4}{x+6}\right)x^2(x+6) + \left(\frac{1}{x^{2}}\right)x^2(x+6) = \frac{x+10}{x^2(x+6)}\cdot x^2(x+6). \] Simplify each term: - The first term: \(\frac{4}{x+6}\times x^2(x+6)=4x^2\). - The second term: \(\frac{1}{x^2}\times x^2(x+6)=x+6\). - The right side becomes: \(x+10\). Thus, we obtain \[ 4x^2+(x+6)=x+10. \] Subtract \(x+10\) from both sides: \[ 4x^2 + x + 6 - x - 10 = 0. \] This simplifies to \[ 4x^2 - 4 = 0. \] Divide through by 4: \[ x^2-1=0. \] Factor the difference of squares: \[ (x-1)(x+1)=0. \] So, the solutions are \[ x=1 \quad \text{or} \quad x=-1. \] Now, check against the restrictions: \(x \ne 0\) and \(x \ne -6\). Both \(1\) and \(-1\) are acceptable. Thus, the statements that are true are: - \(x=1\) is a solution. - \(x=-1\) is a solution. The statements \(x=0\) is a solution and \(x=-6\) is a solution are false.