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A² = b + c

b² = c + a

c² = a + b

a² - b² = b - a => (a - b) (a + b) = b - a

=> a + b = -1 or a = b

1) a + b ≠ - 1 as c² is not negative, if c is a real number.

a = b => c² = 2 a

similarly, b = c => a² = 2 b (as b+c ≠ -1, as a² is non negative)

and c = a => b² = 2 a

so we get a = b = c = 2

b² = c + a

c² = a + b

a² - b² = b - a => (a - b) (a + b) = b - a

=> a + b = -1 or a = b

1) a + b ≠ - 1 as c² is not negative, if c is a real number.

a = b => c² = 2 a

similarly, b = c => a² = 2 b (as b+c ≠ -1, as a² is non negative)

and c = a => b² = 2 a

so we get a = b = c = 2

solve the equation

{bc+b+c+1+ac+c+a+1+ab+a+b+1}/(ab+b+a+1)(c+1)

(a+a+b+b+c+c+ab+bc+ca) /(ab+b+1+a)(c+1)

we know that

a²=b+c , b²=c+a , c²=a+b

{(b+c)+(b+c) + (a+a)}/(ab+1+(b+c))(c+1)

{a²+a²+2a }/{abc+c+c(b+c)+ab+1+(b+c)}

2a(a+1)/abc+c+c(a²)+ab+1+(a²)

2a(a+1)/abc+c+a²c+ab+1+a²

2a(a+1)/abc+ c(a²+1) +a(b+a)+1

2a(a+1)/abc+c²(a+1)+ac²

2a(a+1)/abc+ac²+c²+ac²

2a(a+1)/abc+c²(a+1+a)

2a(a+1)/abc+c²(2a+1)

2a(a+1)/c(ab+c(2a+1)

2a(a+1)/c(ab+2ac+c)

2a(a+1)/c(ab+ac+ac+c)

2a(a+1)/c(a(b+c)+c(a+1)

2a(a+1)/c(a(a²)+c(a+1)

2a(a+1)/c(a³+ac+c) by solving this we get

2a(a+1)/2a(a+1) = 1