Let's solve the given mathematical problem step-by-step.
Given:
[tex]\[
\sqrt[x]{a} = \sqrt[y]{b} = \sqrt[2]{c}
\][/tex]
and
[tex]\[
a \cdot b \cdot c = 1
\][/tex]
We need to prove that [tex]\(x + y + 2 = 0\)[/tex].
First, let's introduce a common variable [tex]\(k\)[/tex] which represents [tex]\(\sqrt[x]{a} = \sqrt[y]{b} = \sqrt[2]{c}\)[/tex]. That is:
[tex]\[
\sqrt[x]{a} = \sqrt[y]{b} = \sqrt[2]{c} = k
\][/tex]
From this, we can express [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] in terms of [tex]\(k\)[/tex]:
[tex]\[
a = k^x, \quad b = k^y, \quad c = k^2
\][/tex]
Now substitute these expressions into the equation [tex]\(a \cdot b \cdot c = 1\)[/tex]:
[tex]\[
k^x \cdot k^y \cdot k^2 = 1
\][/tex]
Combining the exponents, we get:
[tex]\[
k^{x + y + 2} = 1
\][/tex]
Since the base [tex]\(k\)[/tex] must be a positive real number, the only way [tex]\(k^{x + y + 2} = 1\)[/tex] can be true is if the exponent is zero. Therefore:
[tex]\[
x + y + 2 = 0
\][/tex]
Rearranging this equation, we get:
[tex]\[
x + y + 2 - 2 = 0 - 2
\][/tex]
[tex]\[
x + y = -2
\][/tex]
So, we have proven that:
[tex]\[
x + y + 2 = 0
\][/tex]
Thus, [tex]\(\boxed{x + y + 2 = 0}\)[/tex] is indeed correct.
