\( \left. \begin{array} { l } { \frac { ( p ^ { 3 } ) ^ { x - 1 } \cdot p ^ { - 5 } } { \sqrt[ x ] { p ^ { 3 } } } = 1 } \\ { ( x \in N ) } \end{array} \right. \)
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To solve the equation, we start by simplifying the left side. The expression can be rewritten using properties of exponents. The term \((p^3)^{x-1}\) becomes \(p^{3(x-1)}\), so the whole expression transforms into: \[ \frac{p^{3(x-1)} \cdot p^{-5}}{\sqrt[x]{p^3}} = 1. \] Now, let’s express \(\sqrt[x]{p^3}\) using exponents, which gives us \(p^{3/x}\). So the equation becomes: \[ \frac{p^{3(x-1) - 5}}{p^{3/x}} = 1. \] By combining the exponents in the numerator, we have: \[ p^{3(x-1) - 5 - \frac{3}{x}} = 1. \] Since \(p^0 = 1\), we can set the exponent equal to zero: \[ 3(x - 1) - 5 - \frac{3}{x} = 0. \] This simplifies to: \[ 3x - 3 - 5 - \frac{3}{x} = 0 \quad \Rightarrow \quad 3x - 8 - \frac{3}{x} = 0. \] Multiplying through by \(x\) to eliminate the fraction yields: \[ 3x^2 - 8x - 3 = 0. \] Now we can solve this quadratic using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): Here, \(a = 3\), \(b = -8\), and \(c = -3\). Substituting: \[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} = \frac{8 \pm \sqrt{64 + 36}}{6} = \frac{8 \pm \sqrt{100}}{6} = \frac{8 \pm 10}{6}. \] This gives us two potential solutions: 1. \(x = \frac{18}{6} = 3\) 2. \(x = \frac{-2}{6} = -\frac{1}{3}\) (not a natural number) The only valid solution \(x\) in natural numbers is: \[ \boxed{3}. \]
