For `x le 2," solve "x^(3)3^(|x-2|)+3^(x+1) = x^(3)*3^(x-2)+3^(|x-2|+3)`

Explanation

To solve the equation \( x^3 \cdot 3^{|x-2|} + 3^{x+1} = x^3 \cdot 3^{x-2} + 3^{|x-2| + 3} \) for \( x \leq 2 \), we can break it down into two cases based on the value of \( x \).

### Step 1: Analyze the Cases
Since we have the absolute value \( |x-2| \), we need to consider two cases:
1. Case 1: \( x = 2 \)
2. Case 2: \( x < 2 \)

### Step 2: Solve Case 1: \( x = 2 \)
Substituting \( x = 2 \) into the equation:
\[ LHS = 2^3 \cdot 3^{|2-2|} + 3^{2+1} = 8 \cdot 3^0 + 3^3 = 8 \cdot 1 + 27 = 8 + 27 = 35 \]
\[ RHS = 2^3 \cdot 3^{2-2} + 3^{|2-2| + 3} = 8 \cdot 3^0 + 3^{0 + 3} = 8 \cdot 1 + 27 = 8 + 27 = 35 \]
Since \( LHS = RHS \), \( x = 2 \) is a solution.

### Step 3: Solve Case 2: \( x < 2 \)
In this case, \( |x-2| = 2 - x \). The equation becomes:
\[ x^3 \cdot 3^{2-x} + 3^{x+1} = x^3 \cdot 3^{x-2} + 3^{(2-x) + 3} \]
This simplifies to:
\[ x^3 \cdot 3^{2-x} + 3^{x+1} = x^3 \cdot 3^{x-2} + 3^{5-x} \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ x^3 \cdot 3^{2-x} - x^3 \cdot 3^{x-2} = 3^{5-x} - 3^{x+1} \]
Factoring out \( x^3 \):
\[ x^3 (3^{2-x} - 3^{x-2}) = 3^{5-x} - 3^{x+1} \]

### Step 5: Simplifying Further
Notice that \( 3^{2-x} = \frac{3^2}{3^x} = \frac{9}{3^x} \) and \( 3^{x-2} = \frac{3^x}{3^2} = \frac{3^x}{9} \). Thus:
\[ 3^{2-x} - 3^{x-2} = \frac{9 - 3^x}{3^x} = \frac{9 - 3^x}{9} \]
So we have:
\[ x^3 \cdot \frac{9 - 3^x}{9} = 3^{5-x} - 3^{x+1} \]

### Step 6: Solving the Equation
We can now solve this equation for \( x < 2 \). Testing values or using numerical methods may be necessary here.

### Conclusion
From the analysis, we found that \( x = 2 \) is a valid solution. Further investigation into the case \( x < 2 \) may yield additional solutions, but the complexity suggests numerical or graphical methods might be required.