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Question 32
If x + y + z = 6 and $$x^{2}+y^{2}+z^{2}$$=20 then the value of $$x^{3}+y^{3}+z^{3}$$-3xyz is
Solution
We know that $$x^{3}+y^{3}+z^{3}-3xyz = (x + y + z)(x^2 +y^2 + z^2 -xy-yz-xz)$$
$$x^{3}+y^{3}+z^{3}-3xyz = (6)(20 -xy-yz-xz)$$
Hence the solution must be a multiple of 6.
Out of the given options only Option C is a multiple of 6.
Hence Option C is the correct answer.
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