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If (x+y+z=0), what is the value of ( x^3+y^3+z^3-3xyz )?

Answer and explanation

Correct answer: (0)

The direct answer is option D: 0. Use the identity \(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\). The question gives \(x+y+z=0\). Substituting this into the first factor gives \(0\) multiplied by the second factor, so the whole expression is 0. Option D is correct. Option A, \(xyz\), is not the value because the expression contains several cube and product terms, not only \(xyz\). Option B, \(3xyz\), would ignore the cube terms and the given condition. Option C, \(x^2+y^2+z^2\), is only part of the factorisation and is not the whole expression. The important step is to notice the factor \(x+y+z\) before trying separate numerical calculations. Memory cue: whenever three terms satisfy sum zero, this identity makes the expression vanish.

Related tags

Three Variable IdentityZero ConditionCube

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