Answer:
ab+bc+ca=0
Step-by-step explanation:
We are given:
[tex]x^a=y^b=z^c[/tex]
xyz=1
Separating equations:
[tex]x^a=z^c[/tex]
[tex]y^b=z^c[/tex]
Solving the first equation for x:
[tex]\displaystyle x=z^\frac{c}{a}[/tex]
Solving the second equation for y:
[tex]\displaystyle y=z^\frac{c}{b}[/tex]
Substituting in
xyz=1:
[tex]\displaystyle z^\frac{c}{a}z^\frac{c}{b}z=1[/tex]
Adding the exponents:
[tex]\displaystyle z^{\frac{c}{a}+\frac{c}{b}+1}=1[/tex]
Since [tex]1=z^0[/tex]:
[tex]\displaystyle z^{\frac{c}{a}+\frac{c}{b}+1}=z^0[/tex]
The base is z on both sides so we get rid of them:
[tex]\displaystyle \frac{c}{a}+\frac{c}{b}+1=0[/tex]
Multiplying by ab:
bc+ac+ab=0
Reordering:
ab+bc+ca=0
