Question

If $(xy{)}^{(a-1)}=z,(yz{)}^{(b-1)}=x$ and ${\left(xz\right)}^{(c-1)}=y,$ and $xyz$ is not $-1$ or $1$, then prove that $ab+bc+ca=2abc$

Answer 1

We have,

$(xy{)}^{(a-1)}=z$ and $(yz{)}^{(b-1)}=x$ & $(xz{)}^{(c-1)}=y$

Now,

$(xy{)}^{(a-1)}=z$

$log(xy{)}^{(a-1)}=z$

$log(xy{)}^{(a-1)}=logz$ $[\therefore log(m^n)=nlog(m\left)\right]$

$(a-1)log\left(xy\right)=logz$

$a-1=\frac{log\left(z\right)}{log\left(xy\right)}$

$a=1+\frac{log\left(z\right)}{log\left(xy\right)}$

$a=\frac{log\left(xy\right)+log\left(z\right)}{log\left(xy\right)}$

$a=\frac{log\left(xyz\right)}{log\left(xy\right)}\Rightarrow \frac{1}{a}=\frac{log\left(xy\right)}{log\left(xyz\right)}$

similarly,

$\frac{1}{b}=\frac{log\left(yz\right)}{log\left(xyz\right)},\frac{1}{c}=\frac{log\left(xz\right)}{log\left(xyz\right)}$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{log\left(xy\right)}{log\left(xyz\right)}+\frac{log\left(yz\right)}{log\left(xyz\right)}+\frac{log\left(xz\right)}{log\left(xyz\right)}$

$\frac{ab+bc+ca}{abc}=\frac{log\left(xy\right)+log\left(yz\right)+log\left(xz\right)}{log\left(xyz\right)}$

$\frac{ab+bc+ca}{abc}=log\frac{(xy.yz.zx)}{log\left(xyz\right)}$

$\frac{ab+bc+ca}{abc}=\frac{log(xyz{)}^2}{log\left(xyz\right)}\Rightarrow \frac{ab+bc+ca}{abc}=\frac{2log\left(xyz\right)}{log\left(xyz\right)}$

$ab+bc+ca=2abc$