#math problem 08-05-2026 Algebra.

@MathMath901 A little lesson in elegance... a + b = -c b + c = -a a + c = -b ((-c)² + (-a)² + (-b)²) / (a² + b² + c²) (c² + a² + b²) / (a² + b² + c²) = 1 In a zero sum system, any two variables imply a third. C.S. Peirce would call this iconic or abductive reasoning. Very useful.

@MathMath901 Does it work? I made extra step. I get 1 as a final result.

@MathMath901 It could still be undefined

@MathMath901 Numerator=(2a^2+2b^2+2c^2+2ab+2bc+2ac)= (2a^2+2b^2+2c^2+2a(b+c)+2bc)= (2a^2+2b^2+2c^2-2a^2+2bc)= (2b^2+2c^2+2bc)= (2b^2+2c^2+2bc)= (b^2+c^2+(b+c)^2)= a^2+b^2+c^2 So, the quotien is 1

@MathMath901 put a=0 b=1 and c=-1 Numerator = 2 Denominator = 2 Answer = 2/2 =1

@MathMath901 1

@MathMath901 a + b = - c => (a+b)^2 = c^2 (1) b + c = -a => (b+c)^2 = a^2 (2) a + c = -b => (a+c)^2 = b^2 (3) (1) + (2) + (3) => (a+b)^2 + (b+c))^2 + (a+c)^2 = = a^2 + b^2 + c^2 => [(a+b)^2 + (b+c))^2 + (a+c)^2] / (a^2 + b^2 + c^2) = 1

@MathMath901 =1 Took me around 30 secs.

@MathMath901 1

@MathMath901 1

@MathMath901 (a+b)² = (-c)², (b+c)² = (-a)², (a+c)² = (-b)² ---> quotient=1

@MathMath901 a+b=0, c=0 =1

@MathMath901 1