Question

a,b,c,d are natural numbers such that a=bc,b=cd,c=da,d=ab.then (a+b)(b+c)(c+d)(d+a) is equal to

Answer 1

Hey

multiplying all;abcd=1

a/b=bc/cd

a/b=b/d d=ab

a*ab=b^2

a^2=b

By substituting equation ,

d=a^3

c=a^4

b=a^2

a=a

multiplying these all

abcd=a^10

as we have earlier

abcd=1

1=a^10

a=1

b=1

c=1

d=1

solution= 16

ur answer is here

Answer 2

Step-by-step explanation:

Given, a = bc , b = cd , c = da , d = ab

multiplying all, we get

     abcd = bc*cd*da*ab

=> abcd = abcd * abcd

=> abcd = 1

Now,   

      a/b = bc/cd

=> a/b =b/d

=> d=ab

=> a*ab = b2

=> a2 = b

By substituting equations, we get

d=a3

c=a4

b=a2

a=a

multiplying these all

abcd = a10

as we have earlier

abcd=1

     1=a10

=> a=1

=> b=1

=> c=1

=> d=1

Now,  (a+b)*(b+c)*(c+d)*(d+a) = (1+1)*(1+1)*(1+1)*(1+1) = 16

From the option,

(a+b+c+d)2 = (1+1+1+1)2 = 16