To solve the equation:

\[
x^3 + 5x^2 - 7x + 18 = 15
\]

### Step 1: Simplify the equation
First, subtract 15 from both sides to set the equation to 0:

\[
x^3 + 5x^2 - 7x + 18 - 15 = 0
\]

\[
x^3 + 5x^2 - 7x + 3 = 0
\]

So, the equation becomes:

\[
x^3 + 5x^2 - 7x + 3 = 0
\]

### Step 2: Attempt factoring
For a cubic equation like this, factoring directly may be complex. We can try some simple potential rational roots using the **Rational Root Theorem**.

The possible rational roots are the factors of the constant term (3) divided by the factors of the leading coefficient (1). This gives the potential roots:

\[
\pm 1, \pm 3
\]

### Step 3: Test possible roots
Let's test these values by substituting them into the equation \(x^3 + 5x^2 - 7x + 3\).

#### Test \(x = 1\):

\[
1^3 + 5(1^2) - 7(1) + 3 = 1 + 5 - 7 + 3 = 2 \neq 0
\]

So, \(x = 1\) is not a root.

#### Test \(x = -1\):

\[
(-1)^3 + 5(-1)^2 - 7(-1) + 3 = -1 + 5 + 7 + 3 = 14 \neq 0
\]

So, \(x = -1\) is not a root.

#### Test \(x = 3\):

\[
3^3 + 5(3^2) - 7(3) + 3 = 27 + 45 - 21 + 3 = 54 \neq 0
\]

So, \(x = 3\) is not a root.

#### Test \(x = -3\):

\[
(-3)^3 + 5(-3)^2 - 7(-3) + 3 = -27 + 45 + 21 + 3 = 42 \neq 0
\]

So, \(x = -3\) is not a root.

Since simple rational root guesses didn't work, further steps like numerical methods (e.g., Newton's method) or graphing tools are usually employed for more complex cubic equations.

Would you like assistance with these methods, or do you want me to continue solving this step by step?

Learn how to solve the cubic equation x^3 + 5x^2 - 7x + 18 = 15 step by step. This problem covers concepts like the Rational Root Theorem and numerical methods for solving equations. Suitable for high school students in Grades 10-12.

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