Question

a, b, c, d are natural numbers such that a =bc, b=cd,c=da and d=ab. Then (a+b) (b+c) (c + d) (d + a) is equal to
(1)(a + b + c + d)²
(2) (a + b)² + (c + d)²
(3) (a + d)² + (b + c)²
(4) (a + c)² + (6 + d)²
no irrelevant answers !!otherwise I will report your answer!!​

Answer 1

Answer:

(1) (a + b + c + d)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