If a+b+c=0 then prove that a3+b3+c3 =3abc
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Formula:a3+b3+c3-3abc=
=[a+b+c](a2+b2+c2-ab-bc-ca)
Substitute a+b+c=0
=[0](a2+b2+c2-ab-bc-ca)
=0. [because anything multiply by 0=0]
a3+b3+c3-3abc=0
a3+b3+c3=3abc
Hence Proved
=[a+b+c](a2+b2+c2-ab-bc-ca)
Substitute a+b+c=0
=[0](a2+b2+c2-ab-bc-ca)
=0. [because anything multiply by 0=0]
a3+b3+c3-3abc=0
a3+b3+c3=3abc
Hence Proved
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