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

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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}
}
$$
The solutions to the equation $$ x^3 + 1331 = 0 $$ are: - $$ x = -11 $$ - $$ x = \frac{11}{2} \times (1 + i\sqrt{3}) $$ - $$ x = \frac{11}{2} \times (1 - i\sqrt{3}) $$

Solve the cubic equation using factoring and the quadratic formula.

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