Question

=z^c \\xyz=1\\ab+bc+ca=?" alt="x^a=y^b=z^c \\xyz=1\\ab+bc+ca=?" align="absmiddle" class="latex-formula">

Answer 1

Answer:

ab+bc+ca=0

Step-by-step explanation:

We are given:

$x^a=y^b=z^c$

xyz=1

Separating equations:

$x^a=z^c$

$y^b=z^c$

Solving the first equation for x:

$\displaystyle x=z^\frac{c}{a}$

Solving the second equation for y:

$\displaystyle y=z^\frac{c}{b}$

Substituting in

xyz=1:

$\displaystyle z^\frac{c}{a}z^\frac{c}{b}z=1$

Adding the exponents:

$\displaystyle z^{\frac{c}{a}+\frac{c}{b}+1}=1$

Since $1=z^0$:

$\displaystyle z^{\frac{c}{a}+\frac{c}{b}+1}=z^0$

The base is z on both sides so we get rid of them:

$\displaystyle \frac{c}{a}+\frac{c}{b}+1=0$

Multiplying by ab:

bc+ac+ab=0

Reordering:

ab+bc+ca=0