x3−3x2y−xy2+3y3=0
⇒x(x2−y2)−3y(x2−y2)=0
⇒(x−3y)(x−y)(x+y)=0
Now, x=y∀(x,y)∈N×N so its a reflexive relation.
But not symmetric & transitive
Since, (3,1) satisfies but (1,3) does not.
Also (3,1)&(1,-1) satisfies but (3,−1) does not.
Let N be the set of natural numbers and a relation R on N be defined by R=(x,y)∈N×N:x3−3x2y−xy2+3y3=0. Then the relation R is
Held on 27 Jul 2021 · Verified 6 Jul 2026.
symmetric but neither reflexive nor transitive
reflexive but neither symmetric nor transitive
reflexive and symmetric, but not transitive
an equivalence relation
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