If (x,y) (y,z) (z,x) are vertices of a triangle whose centroid is origin prove that x3+y3+z3=3xyz
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Coordinates of Centroid (G) = 0,0 (since coordinates of origin= 0,0)
therefore by formula that is G{x= (x1+x2+x3)/3 & y=(y1+y2+y3)/3}
therefore by putting the values
0=(x+y+z)/3 & 0=(y+z+x)/3
therefore x+y+z=0 from both equations
by alzebric Identity that is x^3+y^3+z^3 = 3xyz (if x+y+z=0)
therefore x^3+y^3+z^3 = 3xyz
HENCE PROVED
therefore by formula that is G{x= (x1+x2+x3)/3 & y=(y1+y2+y3)/3}
therefore by putting the values
0=(x+y+z)/3 & 0=(y+z+x)/3
therefore x+y+z=0 from both equations
by alzebric Identity that is x^3+y^3+z^3 = 3xyz (if x+y+z=0)
therefore x^3+y^3+z^3 = 3xyz
HENCE PROVED
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