Math, asked by Lipimishra2, 1 year ago

a^b = b^c = c^d = d^a

Find bcd + cda + abc + cba, if abcd = 1.

Answers

Answered by Divyankasc
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

Divyankasc: I get 2 solns onnly
abhi178: :-) sorry ,
Divyankasc: For a,b,c,d = 1.. a^b = b^c = c^d = d^a and abcd = 1
abhi178: but am i wrong , or
Divyankasc: Same for -1
abhi178: if you try this with hit and trail i think many possiblilty dear !!!! we will confuse
mysticd: suppose i take a=1 , b=-1, c=1, d=-1 is it correct
Divyankasc: nah
Divyankasc: Because it doesnt fulfil first condition of A^b = b^c = c^d = d^a
mysticd: okk
Answered by abhi178
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

Lipimishra2: Now I know.
Lipimishra2: Thanks anyway.
Divyankasc: Oh my..U are right abhi!
Divyankasc: Well then I can be right too! :P
abhi178: Nah !!! both are right , if questions are different , by the way thanks to both of you
Lipimishra2: Umm. So. Both are right?
Lipimishra2: Or abhi only?
abhi178: dear !!! don't be confuse okay !!! if your questions according to my assumption then i am correct , if divyankasc 's assumption then divankasc are correct . simple :-)
abhi178: here you say , cba not cbd so, in this way i correct
Lipimishra2: Okay.
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