<s>[INST] Given: $\frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y},$ prove: a) $x^x\times y^y\times z^z =1,$ b) $x\times y \times z =1.$ Given: $  \frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z}  = \frac{\log_{10}x}{z-y} $ 
Prove: $   a)\ x^x\times y^y\times  z^z =1 \\ b) \ x\times y \times z =1 $
Here is a few step I tried to solve: $$ \log_{10}z^zy^yx^x => z\log_{10}z + y\log_{10}y + x\log_{10}x  $$ and find each $\log_{10}z, \log_{10}y \ and \log_{10}x$ in the form of one given the equation by equalizing the denominator:
$(x-z)(y-x)\log_{10}z = (y-x)(z-y)\log_{10}y = (y-x)(x-z)log_{10}x $
and calculating  $ \frac{\log_{10}z }{\log_{10}y}, \frac{\log_{10}z }{\log_{10}x} $
Somehow I couldn't figure out the how to get $ x^x\times y^y\times  z^z =1... $
Can you help me to prove the equation? Thank you.
 [/INST] Let $k$ be equal to our fractions:
$$ \frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z}  = \frac{\log_{10}x}{z-y}=k.$$
Thus,  $$\prod_{cyc}z=\prod_{cyc}10^{k(y-x)}=10^0=1.$$
Also, $$\prod_{cyc}z^z=\prod_{cyc}10^{kz(y-x)}=10^0=1.$$
I used $$\sum_{cyc}k(y-x)=k(y-x+z-y+x-z)=0$$ and
$$\sum_{cyc}kz(y-x)=k(zy-zx+xz-xy+yx-yz)=0.$$
</s>