We have,
ax−1=bc
ax=abc
a=(abc)1/x
Similarly,
b=(abc)1/y
c=(abc)1/z
Therefore,
abc=(abc)1/x.(abc)1/y.(abc)1/z
abc=(abc)1/x+1/y+1/z
∴1/x+1/y+1/z=1
xy+yz+zxxyz=1
xy+yz+zx=xyz
xy+yz+zx−xyz=0
Hence, this is the answer.
If ax−1=bc, by−1=ac, cz−1=ab such that x,y,z are integers then value of xy+yz+zx–xyz is