To solve the equation [tex]\( p^{p+2} \cdot \sqrt{x^6} = x^2 \)[/tex] for [tex]\( p \)[/tex], we proceed as follows:
First, rewrite the given equation:
1. Recognize that [tex]\(\sqrt{x^6} = x^{6/2} = x^3\)[/tex].
So the equation becomes:
[tex]\[
p^{p+2} \cdot x^3 = x^2
\][/tex]
2. To isolate the variable [tex]\( p \)[/tex], divide both sides of the equation by [tex]\( x^3 \)[/tex]:
[tex]\[
p^{p+2} = \frac{x^2}{x^3} = x^{2-3} = x^{-1}
\][/tex]
Hence, the equation simplifies to:
[tex]\[
p^{p+2} = x^{-1}
\][/tex]
3. Consider the given options for [tex]\( p \)[/tex]: 0, -1, 1. Evaluate each one to see if they satisfy the equation.
- For [tex]\( p = 0 \)[/tex]:
[tex]\[
0^{0+2} = 0^2 = 0
\][/tex]
Since [tex]\( 0 \neq x^{-1} \)[/tex] for any non-zero [tex]\( x \)[/tex], [tex]\( p = 0 \)[/tex] does not satisfy the equation.
- For [tex]\( p = -1 \)[/tex]:
[tex]\[
(-1)^{-1+2} = (-1)^1 = -1
\][/tex]
Since [tex]\( -1 \neq x^{-1} \)[/tex] for any positive [tex]\( x \)[/tex], [tex]\( p = -1 \)[/tex] does not satisfy the equation.
- For [tex]\( p = 1 \)[/tex]:
[tex]\[
1^{1+2} = 1^3 = 1
\][/tex]
Since [tex]\( 1 \neq x^{-1} \)[/tex] for any [tex]\( x \neq 1 \)[/tex], [tex]\( p = 1 \)[/tex] does not satisfy the equation.
Thus, none of the choices [tex]\( p = 0 \)[/tex], [tex]\( p = -1 \)[/tex], or [tex]\( p = 1 \)[/tex] satisfy the equation [tex]\( p^{p+2} \cdot \sqrt{x^6} = x^2 \)[/tex].
Therefore, the correct answer is:
[tex]\[
\boxed{\text{None of these}}
\][/tex]
