solve....... it plz fast
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a+b+c=0
(a+b+c)³=0
a³+b³+c³-3abc=0
a³+b³+c³=3abc
a³+b³+c³/abc=3
a³/abc + b³/abc + c³/abc=3
Thus, a²/bc + b²/ca + c²/ab=3
Hence proved.
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komal2812:
it is not there
ohh see if u can find it somewhere else
I don't think this proof is correct since there is no formula like (a+b+c) ^3=a^3+b^3+c^3+3abc
no there is a formula
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
this identity is there
yes it is but only when a+b+c =0 and that's the thing which we have to prove, we can't directly use it
no that's given
see the question properly
I know it is given, ok don't discuss it much, we have already given our solutions
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Here's your solution buddy
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