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.


Create a FREE account and get:

  • Free SSC Study Material - 18000 Questions
  • 230+ SSC previous papers with solutions PDF
  • 100+ SSC Online Tests for Free

cracku

Boost your Prep!

Download App