If x^1/a=y^1/b=z=^1/c and xyz=1 let us show that a+b+c=0.
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x^1/a=y^1/b=z^1/c
y=x^b/a & z=x^c/a
x*y*z=x^(1+b/a+c/a) = 1
x^((a+b+c)/a) = 1
1 can be written as x^0
x^0=1
hence
x^((a+b+c)/a) = x^0 = 1
a+b+c=0
y=x^b/a & z=x^c/a
x*y*z=x^(1+b/a+c/a) = 1
x^((a+b+c)/a) = 1
1 can be written as x^0
x^0=1
hence
x^((a+b+c)/a) = x^0 = 1
a+b+c=0
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