Solve for (x, y, z) if x + y + z = 3, x2 + y2 + z2 = 3, x3 + y3 + z3 = 3
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(x+y+z)²=(3)²
x²+y²+z²+2xy+2yz+2zx=9
3+2(xy+yz+zx)=9
2(xy+yz+zx)=9-3
xy+yz+zx =6
x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
3-3xyz=(3)(3-(xy+yz+zx)
3-3xyz=(3)(3-(6)
3-3xyz=3×-3
3-3xyz=-9
-3xyz= -9-3
-3xyz= -12
xyz= -12/-3
xyz= 4
I Hope it's help you
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