Question

6. If x + y + z = 0, show
(i) x + y + z3 = 3xyz ;
(it) (x2 + y2 + xy) = (x2 + x2 + yz) = (22 + x2 + zx).




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Answer 1

Answer:

x+y+z=0 --------1 x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz+zx) --------2 Applying equation 1 in equation 2 x³+y³+z³-3xyz=0 ( x²+y²+z²-xy-yz-zx) x³+y³+z³-3xyz=0 x³+y³+z³=3xyz

Step-by-step explanation:

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