SSC CHSL 12 Oct 2020 Shift 2 Paper

Question : 57 of 100

Marks: +1, -0

x+y+z=10, x^{3}+y^{3}+z^{3}=75

and

x y z=15

, then find the value of

x^{2}+y^{2}+z^{2}-x y-y z-z x

Solution:

We know that,

x^{3}+y^{3}+z^{3}-3 x y z

=(x+y+z)(x^{2}+y^{2}+z^{2}-y z-x y-z x)

75-3 \×15=10 \×(x^{2}+y^{2}+z^{2}-z y-x y-z x)

75-45=10 \×(x^{2}+y^{2}+z^{2}-z y-x y-z x)

30=10 \×(x^{2}+y^{2}+z^{2}-z y-x y-x z)

∴

x^{2}+y^{2}+z^{2}-z y-x y-z x

={30}/{10}=3(\∵x+y+z=10

x^{3}+y^{3}+z^{3}=75, x y z=15)

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