Question

a^2 = b^= c^2 = d^2 find bcd + cda + abc + cba if abcd = 1.

Answer 1

a² = b² = c² = d²

bcd + cda + abc + cba = ?
if abcd = 1

a² = b² = c² = d² = K ( let )

a = ±√k
b = ±√k
c = ±√k
d = ±√k

now,
abcd = 1

(±√k)(±√k)(±√k)(±√k) = 1
k² = 1
K = ± 1
but a = b = c = d = ±√k
K ≥ 0 so, K = 1 and k ≠ - 1

so, a = ±1
b = ± 1
c = ± 1
d = ± 1

now ,
if all are positive e.g a = b = c = d = 1
then,
bcd + cda + abc + cba = 1×1 × 1 + 1 × 1 × 1 + 1 × 1 × 1 + 1 × 1 × 1 = 1 + 1 + 1 + 1 = 4


if all are negative , e.g a = b = c = d = -1
then,
(-1)× (-1)×(-1) + (-1)× (-1) × (-1) + (-1) × (-1) × (-1) + (-1) × (-1) × (-1) = -1 -1 -1 -1 = -4

hence,
value of
bcd + cda + abc + cba = ±4

Answer 2

Given ---
a² = b² = c² = d²
abcd = 1

When you find 3 or more numbers equal then take it as 'k'
a² = b² = c² = d² = k

a² = k
a =± √k

b² = k
b = ±√k

c² = k
c = ±√k

d² = k
d = ±√k

Put the values of a, b , c and d in 'abcd = 1'

(±√k)(±√k)(±√k)(±√k) = 1
k² = 1
k= ± 1

But we found out that  a = b = c = d = ±√k

 k = 1 and k ≠ - 1

so, a = ±1
b = ± 1
c = ± 1
d = ± 1

We have two possible cases

Case - I , When , a = b = c = d = 1

bcd + cda + abc + cba = 1×1 × 1 + 1 × 1 × 1 + 1 × 1 × 1 + 1 × 1 × 1
= 1 + 1 + 1 + 1
 = 4

Case -II , When a = b = c = d = -1

bcd +cda + abc + cba = (-1) x (-1) x(-1) + (-1) x (-1) x (-1) + (-1) x (-1) x (-1) + (-1) x (-1) x (-1)
 = -1 -1 -1 -1
 = -4

Required answer --- bcd + cda + abc + cba = ±4

Hope This Helps You!