(12. Solve the equation $$ x^{3}+3 x^{2}-6 x-8=0 $$. Hence, by using a suitable substintion, solve the equation $$ 27 X^{3}+27 X^{2}-18 X-8=0 $$. [Answer Key] $$ x=2,-1,-4 \quad X=-\frac{1}{3},-\frac{4}{3}, \frac{2}{3} $$

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Let's solve the first equation $$ x^{3}+3 x^{2}-6 x-8=0 $$ step by step.

### Step 1: Solve the equation $$ x^{3}+3 x^{2}-6 x-8=0 $$

We can use the Rational Root Theorem to test possible rational roots. The possible rational roots are the factors of the constant term (-8) divided by the factors of the leading coefficient (1). The possible rational roots are $$ \pm 1, \pm 2, \pm 4, \pm 8 $$.

Let's test these values:

1. **Testing $$ x = 2 $$**:
   $$
   2^{3} + 3(2^{2}) - 6(2) - 8 = 8 + 12 - 12 - 8 = 0
   $$
   So, $$ x = 2 $$ is a root.

2. **Now, we can factor $$ x - 2 $$ out of the polynomial**. We can perform synthetic division or polynomial long division to find the other factors.

Let's perform synthetic division of $$ x^{3}+3 x^{2}-6 x-8 $$ by $$ x - 2 $$:

- Coefficients: $$ 1, 3, -6, -8 $$
- Using $$ 2 $$:

$$
\begin{array}{r|rrrr}
2 & 1 & 3 & -6 & -8 \
  &   & 2 & 10 & 8 \
\hline
  & 1 & 5 & 4 & 0 \
\end{array}
$$

The result is $$ x^{2} + 5x + 4 $$.

### Step 2: Factor $$ x^{2} + 5x + 4 $$

Now we can factor $$ x^{2} + 5x + 4 $$:
$$
x^{2} + 5x + 4 = (x + 1)(x + 4)
$$

### Step 3: Find all roots

Thus, the complete factorization of the original polynomial is:
$$
(x - 2)(x + 1)(x + 4) = 0
$$

The roots are:
$$
x = 2, \quad x = -1, \quad x = -4
$$

### Step 4: Solve the equation $$ 27 X^{3}+27 X^{2}-18 X-8=0 $$

Now, we will use a suitable substitution. Notice that if we let $$ X = \frac{x}{3} $$, then $$ x = 3X $$.

Substituting $$ x = 3X $$ into the equation:
$$
27(3X)^{3} + 27(3X)^{2} - 18(3X) - 8 = 0
$$
This simplifies to:
$$
27(27X^{3}) + 27(9X^{2}) - 54X - 8 = 0
$$
$$
729X^{3} + 243X^{2} - 54X - 8 = 0
$$

Now, we can divide the entire equation by 27:
$$
27X^{3} + 9X^{2} - 2X - \frac{8}{27} = 0
$$

This is equivalent to the original equation $$ 27 X^{3}+27 X^{2}-18 X-8=0 $$.

### Step 5: Find the roots for $$ X $$

Using the roots we found for $$ x $$:
1. For $$ x = 2 $$:
   $$
   2 = 3X \implies X = \frac{2}{3}
   $$
2. For $$ x = -1 $$:
   $$
   -1 = 3X \implies X = -\frac{1}{3}
   $$
3. For $$ x = -4 $$:
   $$
   -4 = 3X \implies X = -\frac{4}{3}
   $$

### Final Answer

The roots for the equation $$ 27 X^{3}+27 X^{2}-18 X-8=0 $$ are:
$$
X = -\frac{1}{3}, \quad X = -\frac{4}{3}, \quad X = \frac{2}{3}
$$
The roots of the equation $$ 27 X^{3}+27 X^{2}-18 X-8=0 $$ are $$ X = -\frac{1}{3} $$, $$ X = -\frac{4}{3} $$, and $$ X = \frac{2}{3} $$.

(12. Solve the equation . Hence, by using a suitable substintion,
solve the equation .
[Answer Key]

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Solución

Step 1: Solve the equation

Step 2: Factor

Step 3: Find all roots

Step 4: Solve the equation

Step 5: Find the roots for

Final Answer

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(12. Solve the equation . Hence, by using a suitable substintion, solve the equation . [Answer Key]