If x+y+z=7, x^{2}+y^{2}+z^{2}=85 and x^{3}+y^{3}+z^{3}=913, then the value of \sqrt[3]{x y z} is

SSC CGL 17 Aug 2021 Shift 1 Paper

Question : 66 of 100

Marks: +1, -0

x+y+z=7, x^{2}+y^{2}+z^{2}=85

and

x^{3}+y^{3}+z^{3}=913

, then the value of

\;^{3}√{x y z}

Solution:

Given,

x+y+z=7

x^{2}+y^{2}+z^{2}=85

and

x^{3}+y^{3}+z^{3}=913

We know that,

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

+2(x y+y z+z x)

⇒(7)^{2}=85+2(x y+y z+z x)

⇒ 2(x y+y z+z x)=49-85=-36

⇒ x y+y z+z x=-18

Now,

x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)

\;\;\; (x^{2}+y^{2}+z^{2}-x y-y z-z x)

⇒ 913-3 x y z=7(85+18)

⇒ 3 x y z=913-7 × 103=913-721

⇒ \;\; x y z=\;{192}/{3}=64

∴ \;\; \;^{3}√{x y z}=\;^{3}√{64}=4

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