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(x+y)²/ xy + (y+z)1/yz (z+x)1/xz if x+y+z = 0 |
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Answer» -STEP explanation:Givenx+y+z=1,xy+yz+zx=−1 and xyz=−1X 3 +y 3 +z 3 −3xyz=(x+y+z)(x 2 +y 2 +z 2 −xy−yz−zx)x 3 +y 3 +z 3 −3(−1)=1(x 2 +y 2 +z 2 )−(−1)x 3 +y 3 +z 3 +3=x 2 +y 2 +z 2 +1x 3 +y 3 +z 3 +2=x 2 +y 2 +z 2 (x+y+z) 2 =x 2 +y 2 +z 2 +2xy+2yz+2xz1 2 =x 2 +y 2 +z 2 +2(−1)x 2 +y 2 +z 2 =1+2x 2 +y 2 +z 2 =3x 3 +y 3 +z 3 +2=3x 3 +y 3 +z 3 =1 |
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