If a³+b³+c³ = 3abc ,then find the value (a+b+c)
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we know the identity,
x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-xz)
from that
if a³+b³+c³=3abc
then
3abc-3abc=(a+b+c)(a²+b²+c²-ab-bc-ac)
0=(a+b+c)(a²+b²+c²-ab-bc-ac)
⇒(a+b+c)=0 or (a²+b²+c²-ab-bc-ac)=0
x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-xz)
from that
if a³+b³+c³=3abc
then
3abc-3abc=(a+b+c)(a²+b²+c²-ab-bc-ac)
0=(a+b+c)(a²+b²+c²-ab-bc-ac)
⇒(a+b+c)=0 or (a²+b²+c²-ab-bc-ac)=0
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