x+y+z=0 show that x cube +y cube +z cube = 3xyz
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Answer:
x³ + y³ +z³ =3xyz
Step-by-step explanation:
in the context of the given question ;
we have to find prove that x³ + y³ +z³ = 3xyz if x+y+z = 0
given ;
x+y+z = 0
we have to prove that;
x³ + y³ +z³ = 3xyz
we know that the identity if
x³ + y³ +z³ -3xyz =(x+y+z)(x²+y²+z²-xy-yz-zx)
but we already know that
(x+y+z=0)
so therefore;
x³ + y³ +z³ -3xyz =(0)(x²+y²+z²-xy-yz-zx)
x³ + y³ +z³ -3xyz =0
therefore,
x³ + y³ +z³ =3xyz
∴ Hence proved
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