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Algebra (Olympiad): Nonzero Real Numbers Satisfy System Find Value Expression

JEE Maths question with a full step-by-step solution.

Question
The nonzero real numbers a,b,ca, b, c satisfy the system
a2+a=b2,b2+b=c2,c2+c=a2.a^2 + a = b^2, \qquad b^2 + b = c^2, \qquad c^2 + c = a^2.
 find the value of the expression (a−b)(b−c)(c−a)(a-b)(b-c)(c-a).
Solution
Answer: 1
Step 1: Add the three equations.
(a2+a)+(b2+b)+(c2+c)=b2+c2+a2.(a^2 + a) + (b^2 + b) + (c^2 + c) = b^2 + c^2 + a^2.
 The squared terms cancel, leaving
a+b+c=0....(i)a + b + c = 0. \qquad \text{...(i)}
 Step 2: Rewrite each equation as a difference of squares.
a2−b2=−a...(ii),b2−c2=−b...(iii),c2−a2=−c...(iv).a^2 - b^2 = -a \quad \text{...(ii)}, \qquad b^2 - c^2 = -b \quad \text{...(iii)}, \qquad c^2 - a^2 = -c \quad \text{...(iv)}.
 Step 3: Multiply (ii), (iii) and (iv).
(a2−b2)(b2−c2)(c2−a2)=(−a)(−b)(−c)=−abc.(a^2 - b^2)(b^2 - c^2)(c^2 - a^2) = (-a)(-b)(-c) = -abc.
 Factoring each difference of squares on the left,
(a−b)(a+b) (b−c)(b+c) (c−a)(c+a)=−abc.(a-b)(a+b)\,(b-c)(b+c)\,(c-a)(c+a) = -abc.
 Step 4: Apply (i). From a+b+c=0a + b + c = 0 we get a+b=−ca + b = -c, b+c=−ab + c = -a and c+a=−bc + a = -b. Substituting,
(a−b)(b−c)(c−a) (−c)(−a)(−b)=−abc,(a-b)(b-c)(c-a)\,(-c)(-a)(-b) = -abc,
 that is
(a−b)(b−c)(c−a) (−abc)=−abc.(a-b)(b-c)(c-a)\,(-abc) = -abc.
 Step 5: Since a,b,ca, b, c are nonzero, abc≠0abc \neq 0. Dividing both sides by −abc-abc,
(a−b)(b−c)(c−a)=1.(a-b)(b-c)(c-a) = 1.
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