To solve this problem, we will start by interpreting the given conditions and transforming them into mathematical equations for manipulation.
Given:
[tex]\[ \sqrt[x]{a} = \sqrt[y]{b} = \sqrt[3]{c} \][/tex]
and
[tex]\[ a \cdot b \cdot c = 1. \][/tex]
We will denote the common value of [tex]\(\sqrt[x]{a}\)[/tex], [tex]\(\sqrt[y]{b}\)[/tex], and [tex]\(\sqrt[3]{c}\)[/tex] by some constant [tex]\(k\)[/tex]. So we can write:
[tex]\[ a^{1/x} = b^{1/y} = c^{1/3} = k. \][/tex]
From this, we can express [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] in terms of [tex]\(k\)[/tex]:
[tex]\[ a = k^x, \][/tex]
[tex]\[ b = k^y, \][/tex]
[tex]\[ c = k^3. \][/tex]
Now, substituting these expressions into the second condition [tex]\( a \cdot b \cdot c = 1 \)[/tex]:
[tex]\[ (k^x) \cdot (k^y) \cdot (k^3) = 1. \][/tex]
Simplify the left side:
[tex]\[ k^x \cdot k^y \cdot k^3 = k^{x+y+3}. \][/tex]
Therefore, the equation becomes:
[tex]\[ k^{x+y+3} = 1. \][/tex]
For [tex]\(k^{x+y+3}\)[/tex] to equal 1, the exponent must be zero because the only way [tex]\(k\)[/tex] raised to any power equals 1 is if the exponent itself is zero (assuming [tex]\(k\)[/tex] is not zero):
[tex]\[ x + y + 3 = 0. \][/tex]
Thus, we have:
[tex]\[ x + y + 3 = 0. \][/tex]
Rewriting this, we find:
[tex]\[ x + y + (-3) = 0, \][/tex]
which confirms that:
[tex]\[ x + y - 3 = 0. \][/tex]
Therefore, the equation we set out to prove is:
[tex]\[ x + y + 3 = 0 \][/tex]
or equivalently:
[tex]\[ x + y + z = 0, \][/tex]
where [tex]\(z = -3\)[/tex].
The final result confirms the given condition [tex]\( x + y + z = 0 \)[/tex].
