An algebra problem by Aman Dubey

Algebra Level 2

Simplify the expression: ( a − b ) 3 + ( b − c ) 3 + ( c − a ) 3 ( a − b ) ( b − c ) ( c − a ) \dfrac{ (a-b)^3 +(b-c)^3 + (c-a)^3}{(a-b)(b-c)(c-a)} .

Details And Assumptions :

  • a ≠ b ≠ c a\neq b\neq c
  • Use :- x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − x z ) x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy -yz - xz)


The answer is 3.

Aman Dubey by Aman Dubey , 22, India

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1 solution

Let a − b = x , b − c = y , c − a = z a-b=x,b-c=y,c-a=z Then x + y + z = 0 x+y+z=0 .Substituting this in the identity: x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − x z − y z ) x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz) We get that x 3 + y 3 + z 3 = 3 x y z x^3+y^3+z^3=3xyz .Hence: x 3 + y 3 + z 3 x y z = 3 x y z x y z = 3 \frac{x^3+y^3+z^3}{xyz}=\frac{3xyz}{xyz}=\boxed{3}

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