Question

`x^a=y^b=z^c \\xyz=1\\ab+bc+ca=?`

Answer 1

ab+bc+ca=0

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^(c)/(a)`

Solving the second equation for y:

`\displaystyle y=z^(c)/(b)`

Substituting in

xyz=1:

`\displaystyle z^(c)/(a)z^(c)/(b)z=1`

Adding the exponents:

`\displaystyle z^{(c)/(a)+(c)/(b)+1}=1`

Since
`1=z^0`:

`\displaystyle z^{(c)/(a)+(c)/(b)+1}=z^0`

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

`\displaystyle (c)/(a)+(c)/(b)+1=0`

Multiplying by ab:

bc+ac+ab=0

Reordering:

ab+bc+ca=0