## Solving the Equation: (x+3)^4 + (x+5)^4 = 16

This equation might look intimidating at first glance, but we can solve it using a combination of algebraic manipulation and a bit of ingenuity. Here's how we can approach it:

### 1. Expanding and Simplifying

Let's start by expanding the terms:

- (x+3)^4 = x^4 + 12x^3 + 54x^2 + 108x + 81
- (x+5)^4 = x^4 + 20x^3 + 150x^2 + 500x + 625

Now, substituting these back into the original equation:

x^4 + 12x^3 + 54x^2 + 108x + 81 + x^4 + 20x^3 + 150x^2 + 500x + 625 = 16

Combining like terms, we get:

2x^4 + 32x^3 + 204x^2 + 608x + 700 = 16

Subtracting 16 from both sides:

2x^4 + 32x^3 + 204x^2 + 608x + 684 = 0

### 2. Making a Substitution

To simplify this quartic equation, we can make a substitution. Let y = x + 4. This implies x = y - 4. Substituting this into the equation:

2(y-4)^4 + 32(y-4)^3 + 204(y-4)^2 + 608(y-4) + 684 = 0

Expanding and simplifying this will still result in a quartic equation, but it will be easier to work with:

2y^4 - 32y^3 + 128y^2 - 128y = 0

### 3. Solving the Simplified Equation

We can factor out a 2y from the simplified equation:

2y(y^3 - 16y^2 + 64y - 64) = 0

This gives us two possibilities:

**2y = 0**=> y = 0**y^3 - 16y^2 + 64y - 64 = 0**

To solve the cubic equation, we can try to find a rational root using the Rational Root Theorem. However, in this case, there are no rational roots. We would need to use numerical methods (like graphing calculators or numerical solvers) to find approximate solutions.

### 4. Back to the Original Variable

Once we have the solutions for y, we can substitute back to find the solutions for x:

**If y = 0, then x = y - 4 = -4**

The other solutions for y would need to be substituted back to find the corresponding x values.

### 5. Verifying Solutions

Finally, we need to verify if our solutions satisfy the original equation. Substitute each solution for x back into (x+3)^4 + (x+5)^4 = 16 and see if the equation holds true.

**Important Note:** The original equation is a quartic equation, meaning it has up to four solutions. We have found one solution (x = -4), and the remaining solutions might be real or complex numbers.

By using a combination of algebraic manipulation, substitution, and potentially numerical methods, we can solve the seemingly complex equation (x+3)^4 + (x+5)^4 = 16.