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