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Answer :
We know that,
\(x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\)
by using identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
As it is given that (x + y + z = 0)
\(x^3 + y^3 + z^3 - 3xyz = (0)(x^2 + y^2 + z^2 - xy - yz - zx)\)
\(\Rightarrow x^3 + y^3 + z^3 = 3xyz\)
Hence, proved.