if n+y+z=1,ny+yz+zn=1 then Find n³+y³+2³

plz answer very urgent!!

# if n+y+z=1,ny+yz+zn=1 then Find n³+y³+2³

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Answer:Identity:x3 +y3 +z3 -3xyz = (x+y+z)( x2 +y2 +z2 -xy-yz-zx)

Since x + y + z = 1, xy + yz + zx = -1and xyz= -1,

Putting values, we get:

So

Now

(x+y+z)2 = x2 +y2 +z2 +2xy +2yz +2zx

12 = x2 +y2 +z2 +2(-1)

x2 +y2 +z2 = 3

Put in eq (i)

x3 +y3 +z3 =3-2 = 1

Answered by | 4th Jun, 2014, 03:23: PM

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