Question

If x³ + y3 + z³ = 6xyz and xyz = X + y + z, then the value
of (x - y)² + (y – z)² + (z – x)² is equal to
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Answer 1

Answer:

Step-by-step explanation: I think you made a typing error and the first equation must be X^3 + Y^3+Z^3 = 6XYZ. This itself is trivial solution.

A similar correction in the final line be Squares ; i.e (X-Y)^2 + (Y-Z)^2 +(Z-X)^2

Here there are three unknown variables x,y,z. Only two equations are given. hence the solution is ON trivial OR trial and error base only

X+Y+Z = XYZ is matched by 2 sets

X=1; Y =1 and Z =1

X = 1 ; Y=2 ; and Z = 3 [ These three may be interchanged also ]

if X=1; Y =1 and Z =1 ; then it will NOT match X^3 +Y^3+Z^3 = 6XYZ

Therefore, the second possibility X = 1 ; Y=2 ; and Z = 3 matches

SUBSTITUTING this in 3rd equation becomes

(1–2)^2+(2–3)^2+(3–1)^2 = 1+1+4 = 6;