\( 1 \leftarrow \quad \) Solve the cubic equation using factoring and the quadratic formula. \( x^{3}+1331=0 \)

We start with the equation \[ x^3 + 1331 = 0. \] Recognize that \(1331 = 11^3\). Thus, we can rewrite the equation as \[ x^3 + 11^3 = 0. \] This is a sum of cubes, and we use the sum of cubes factorization formula: \[ a^3 + b^3 = (a+b)(a^2-ab+b^2), \] with \(a = x\) and \(b = 11\). Therefore, \[ x^3 + 11^3 = (x+11)(x^2-11x+121)=0. \] The equation factors into two parts: 1. \(x+11=0\) 2. \(x^2-11x+121=0\) --- **1. Solve \(x+11=0\):** \[ x = -11. \] --- **2. Solve \(x^2-11x+121=0\) using the quadratic formula:** The quadratic formula is given by \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, \] where for the equation \(ax^2 + bx + c = 0\), we have \(a=1\), \(b=-11\), and \(c=121\). First, compute the discriminant: \[ \Delta = b^2-4ac = (-11)^2-4(1)(121) = 121-484 = -363. \] Since the discriminant is negative, the solutions will be complex. Now, apply the formula: \[ x=\frac{-(-11)\pm\sqrt{-363}}{2(1)}=\frac{11\pm\sqrt{-363}}{2}. \] Write the square root of a negative number as: \[ \sqrt{-363}=i\sqrt{363}. \] Notice that \(363=121\cdot3\), so \[ \sqrt{363}=\sqrt{121\cdot3}=11\sqrt{3}. \] Thus, the complex solutions are: \[ x=\frac{11\pm i11\sqrt{3}}{2}=\frac{11}{2}\left(1\pm i\sqrt{3}\right). \] --- **Final solutions:** \[ \boxed{ \begin{aligned} x &= -11,\\[1mm] x &= \frac{11}{2}\left(1+ i\sqrt{3}\right),\\[1mm] x &= \frac{11}{2}\left(1- i\sqrt{3}\right). \end{aligned} } \]