Question

If ax - 1 = bc, by -1 = ac, cz - 1 = ab such that x, y, z  integer then value of xy + yz + zx – xyz is​

Answer 1

given, a^(x - 1) = bc

b^(y - 1) = ca

c^(z - 1) = ab

we have to find out the value of xy + yz + zx - xyz

first of all, resolve, a^(x - 1) = bc ,b^(y - 1) = ca and c^(z - 1) = ab

a^(x - 1) = bc

⇒a^x . a^-1 = bc

⇒a^x/a = bc

⇒a^x = abc

⇒a = (abc)^{1/x}

similarly, b^(y - 1) = ca ⇒b = (abc)^{1/y}

and c^(z - 1) = ab ⇒c = (abc)^{1/z}

now, a.b.c = (abc)^{1/x}.(abc)^{1/y}. (abc)^{1/x}

taking log both sides,

log(abc) = log[ (abc)^{1/x}.(abc)^{1/y}. (abc)^{1/x}]

⇒log(abc) = log[(abc)^{1/x + 1/y + 1/z}]

⇒log(abc) = (1/x + 1/y + 1/z) log(abc)

⇒1 = 1/x + 1/y + 1/z

⇒1 = (yz + zx + xy)/xyz

⇒xyz = yz + zx + xy

⇒xy + yz + zx - xyz = 0 [Ans]

Answer 2

Answer:

given, a^(x - 1) = bc

b^(y - 1) = ca

c^(z - 1) = ab

we have to find out the value of xy + yz + zx - xyz

first of all, resolve, a^(x - 1) = bc ,b^(y - 1) = ca and c^(z - 1) = ab

a^(x - 1) = bc

⇒a^x . a^-1 = bc

⇒a^x/a = bc

⇒a^x = abc

⇒a = (abc)^{1/x}

similarly, b^(y - 1) = ca ⇒b = (abc)^{1/y}

and c^(z - 1) = ab ⇒c = (abc)^{1/z}

now, a.b.c = (abc)^{1/x}.(abc)^{1/y}. (abc)^{1/x}

taking log both sides,

log(abc) = log[ (abc)^{1/x}.(abc)^{1/y}. (abc)^{1/x}]

⇒log(abc) = log[(abc)^{1/x + 1/y + 1/z}]

⇒log(abc) = (1/x + 1/y + 1/z) log(abc)

⇒1 = 1/x + 1/y + 1/z

⇒1 = (yz + zx + xy)/xyz

⇒xyz = yz + zx + xy

⇒xy + yz + zx - xyz = 0 [Ans]

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