a^b = b^c = c^d = d^a
Find bcd + cda + abc + cba, if abcd = 1.
Answers
Answered by
4
There are many ways to do it. But, since you are asked the direct answer rather than a solution here, you can directly do it.
Let's do it!
a^b = b^c = c^d = d^a
abcd = 1
Now, think of a value that substitutes a,b,c and d and fulfills the above equations. That is pretty simple.
a, b, c, d = ±1
bcd + cda + abc + cba = ±1 ± 1 ± 1 ± 1 = ± 4
If you are asked for direct answers, use short methods to reach the answer. That is because sometimes real solutions spin minds :D
Let's do it!
a^b = b^c = c^d = d^a
abcd = 1
Now, think of a value that substitutes a,b,c and d and fulfills the above equations. That is pretty simple.
a, b, c, d = ±1
bcd + cda + abc + cba = ±1 ± 1 ± 1 ± 1 = ± 4
If you are asked for direct answers, use short methods to reach the answer. That is because sometimes real solutions spin minds :D
Divyankasc:
I get 2 solns onnly
Answered by
3
a^b = b^c =c^d = d^a = K
a = k^1/b
b = k^1/c
c= k^1/d
d =k^1/a
now,
abcd = 1
(K^1/b)(k^1/c)(k^1/d)(k^1/a) =1
K^(cda+ dab + abc + adc)/abcd =1
But we have given that , abcd = 1 .
We also know that ,
a^0 = 1
If we use this
and
cda + abd+ abc + bcd = 0
then ,
k^0 = 1
So,
bcd +cda + abc + abd = 0
a = k^1/b
b = k^1/c
c= k^1/d
d =k^1/a
now,
abcd = 1
(K^1/b)(k^1/c)(k^1/d)(k^1/a) =1
K^(cda+ dab + abc + adc)/abcd =1
But we have given that , abcd = 1 .
We also know that ,
a^0 = 1
If we use this
and
cda + abd+ abc + bcd = 0
then ,
k^0 = 1
So,
bcd +cda + abc + abd = 0
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