If
/
+
/
= −1 then find
Answers
Answer:
Answer:
Given an equation: a+\frac{1}{b}=3a+
b
1
=3 , b+\frac{1}{c}=4b+
c
1
=4 and c+\frac{1}{a} =\frac{9}{11}c+
a
1
=
11
9
.
Sum of these equation is, a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=7+\frac{9}{11} =\frac{86}{11}a+b+c+
a
1
+
b
1
+
c
1
=7+
11
9
=
11
86
.
Product of the above equation: (a+\frac{1}{b})\cdot (b+\frac{1}{c})\cdot (c+\frac{1}{a})(a+
b
1
)⋅(b+
c
1
)⋅(c+
a
1
)
On Simplify:
abc+a+c+\frac{1}{b} +b+\frac{1}{c}+\frac{1}{a} +\frac{1}{abc} = \frac{108}{11}abc+a+c+
b
1
+b+
c
1
+
a
1
+
abc
1
=
11
108
or
abc+(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(a+b+c+
a
1
+
b
1
+
c
1
) +\frac{1}{abc}
abc
1
= \frac{108}{11}
11
108
.
abc+\frac{86}{11}+\frac{1}{abc}=\frac{108}{11}abc+
11
86
+
abc
1
=
11
108
abc+\frac{1}{abc} = \frac{108}{11}-\frac{86}{11} = \frac{108-86}{11} =\frac{22}{11}abc+
abc
1
=
11
108
−
11
86
=
11
108−86
=
11
22
⇒ abc+\frac{1}{abc} =2abc+
abc
1
=2 .
Let abc be x
then; x+\frac{1}{x} = 2x+
x
1
=2 or
x^2-2x+1=0x
2
−2x+1=0
On solving the quadratic equation; we have,
x=1 or
a\times b \times c=1a×b×c=1