if x^a = y^b = z^c & ab+ bc+ ac = 0 , prove that xyz = 1
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Answered by
0
Answer:
Answer
We have,
a
x−1
=bc
a
a
x
=abc
a
x
=abc
a=(abc)
1/x
…(1)
Similarly,
b=(abc)
1/y
…(2)
c=(abc)
1/z
…(3)
Multiply (1),(2),(3)
abc=(abc)
1/x
.(abc)
1/y
.(abc)
1/z
abc=(abc)
1/x+1/y+1/z
∴
x
1
+
y
1
+
z
1
=1
xyz
xy+yz+zx
=1
xy+yz+zx=xyz
xy+yz+zx−xyz=0
Answered by
1
Answer:
This is an interesting problem. Here is one observation.
a^x = b^y = c^z
Hence
a = c^z/x and b = c^z/y
Hence ab = (c^1/x.c^1/y)^z
Hence ab = c^z(1/x + 1/y) = c^(z(x + y)/xy) = c^(z^2)(x + y)
Similarly bc = a^(x^2)(y + z) and ca = b^(y^2)(z + x)
Hence ab + bc + ca = c^(z^2)(x + y) + a^(x^2)(y + z) + b^(y^2)(z + x)
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