Question

If xyz = 1, then show that ${\left( {1 + x + {y^{ - 1}}} \right)^{ - 1}} + {\left( {1 + y + {z^{ - 1}}} \right)^{ - 1}} + {\left( {1 + z + {x^{ - 1}}} \right)^{ - 1}} = 1$

Answer 1

Hint: In this question use the given information to simplify the given equation and also remember to use the given equation xyz = 1 to find the value of y in terms of x and z, then put it in the LHS, using this information will help you to approach towards the solution of the question.

Complete step-by-step answer:
According to the given information we have xyz = 1 and we have to show that ${\left( {1 + x + {y^{ - 1}}} \right)^{ - 1}} + {\left( {1 + y + {z^{ - 1}}} \right)^{ - 1}} + {\left( {1 + z + {x^{ - 1}}} \right)^{ - 1}}$ is equal to 1
So taking equation ${\left( {1 + x + {y^{ - 1}}} \right)^{ - 1}} + {\left( {1 + y + {z^{ - 1}}} \right)^{ - 1}} + {\left( {1 + z + {x^{ - 1}}} \right)^{ - 1}}$ as L.H.S and 1 as R.H.S
Let’s simplify L.H.S
L.H.S = ${\left( {1 + x + {y^{ - 1}}} \right)^{ - 1}} + {\left( {1 + y + {z^{ - 1}}} \right)^{ - 1}} + {\left( {1 + z + {x^{ - 1}}} \right)^{ - 1}}$
$ \Rightarrow $L.H.S = ${\left( {1 + x + {y^{ - 1}}} \right)^{ - 1}} + {\left( {1 + y + {z^{ - 1}}} \right)^{ - 1}} + {\left( {1 + z + {x^{ - 1}}} \right)^{ - 1}}$
$ \Rightarrow $L.H.S = \[\dfrac{1}{{1 + x + {y^{ - 1}}}} + \dfrac{1}{{1 + y + {z^{ - 1}}}} + \dfrac{1}{{1 + z + {x^{ - 1}}}}\] (equation 1)
Since it is given that xyz = 1 which can be written as \[xz = {y^{ - 1}}\] and \[y = {\left( {xz} \right)^{ - 1}}\]
Substituting these values in the equation 1 we get
$ \Rightarrow $L.H.S = \[\dfrac{1}{{1 + x + xz}} + \dfrac{1}{{1 + {{\left( {xz} \right)}^{ - 1}} + {z^{ - 1}}}} + \dfrac{1}{{1 + z + {x^{ - 1}}}}\]
$ \Rightarrow $L.H.S = \[\dfrac{1}{{1 + x + xz}} + \dfrac{1}{{1 + \dfrac{1}{{xz}} + \dfrac{1}{z}}} + \dfrac{1}{{1 + z + \dfrac{1}{x}}}\]
$ \Rightarrow $L.H.S = \[\dfrac{1}{{1 + x + xz}} + \dfrac{1}{{\dfrac{{xz + 1 + x}}{{xz}}}} + \dfrac{1}{{\dfrac{{x + xz + 1}}{x}}}\]
$ \Rightarrow $L.H.S = \[\dfrac{1}{{1 + x + xz}} + \dfrac{{xz}}{{xz + 1 + x}} + \dfrac{x}{{x + xz + 1}}\]
Since the denominator of all the fractional equation is same therefore
L.H.S = \[\dfrac{{1 + x + xz}}{{1 + x + xz}}\]
So as we can see that numerator and denominator are same therefore
L.H.S = 1
Therefore L.H.S = R.H.S
Hence ${\left( {1 + x + {y^{ - 1}}} \right)^{ - 1}} + {\left( {1 + y + {z^{ - 1}}} \right)^{ - 1}} + {\left( {1 + z + {x^{ - 1}}} \right)^{ - 1}}$ is equal to 1.

Note: The above question was a very basic question which was based on the substitution method we used the given information xyz = 1 to find the value of ${y^{ - 1}}$ and y then we substituted the values in the equation and simplified it until the denominator became the same and then we used the addition operation and got the result 1.