\( \left. \begin{array} { l } { \frac { ( p ^ { 3 } ) ^ { x - 1 } \cdot p ^ { - 5 } } { \sqrt[ x ] { p ^ { 3 } } } = 1 } \\ { ( x \in N ) } \end{array} \right. \)

To solve the equation: \[ \frac{(p^3)^{x - 1} \cdot p^{-5}}{\sqrt[x]{p^3}} = 1 \quad \text{where} \quad x \in \mathbb{N} \] **Step 1: Simplify the Expression** First, simplify each part of the equation using exponent rules. 1. **Numerator:** \[ (p^3)^{x - 1} \cdot p^{-5} = p^{3(x - 1)} \cdot p^{-5} = p^{3x - 3 - 5} = p^{3x - 8} \] 2. **Denominator:** \[ \sqrt[x]{p^3} = (p^3)^{1/x} = p^{3/x} \] 3. **Combine Numerator and Denominator:** \[ \frac{p^{3x - 8}}{p^{3/x}} = p^{3x - 8 - \frac{3}{x}} \] **Step 2: Set the Exponent to Zero** For the equation to hold true (\( p^{\text{exponent}} = 1 \)), the exponent must be zero (assuming \( p \neq 1 \)). \[ 3x - 8 - \frac{3}{x} = 0 \] **Step 3: Solve for \( x \)** Multiply both sides by \( x \) to eliminate the fraction: \[ 3x^2 - 8x - 3 = 0 \] This is a quadratic equation in standard form: \[ 3x^2 - 8x - 3 = 0 \] **Step 4: Use the Quadratic Formula** The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -8 \), and \( c = -3 \). \[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} = \frac{8 \pm \sqrt{64 + 36}}{6} = \frac{8 \pm \sqrt{100}}{6} = \frac{8 \pm 10}{6} \] This gives two potential solutions: 1. \( x = \frac{18}{6} = 3 \) 2. \( x = \frac{-2}{6} = -\frac{1}{3} \) **Step 5: Consider the Domain** Since \( x \in \mathbb{N} \) (natural numbers), the only valid solution is: \[ x = 3 \] **Conclusion** The natural number solution to the equation is: **Answer:** \(x = 3\)