Question

if x+y+z=6 and x square+y square+z square =18,then x cube+y cube+z cube-3xyz
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Answer 1

Answer:

54

Step-by-step explanation:

given: x+y+z=6

x^2+y^2+z^2=18

to find:x^3+y^3+z^3-3xyz

solution: =(x+y+z)(x^2+y^2+z^2-xy-yz-zx)

=6(18-xy-yz-zx)

=6(18-1(xy+yz+zx)) ------(1)

now, (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)

=> 6^2=18+2(xy+yz+zx)

=> 36-18=2(xy+yz+zx)

=> 18=2(xy+yz+zx)

=> (xy+yz+zx)=9 -----(2)

by substituting equation (2) in (1), we get

x^3+y^3+z^3-3xyz=6•(18-1(9))

=> " " " " " = 6•(18-9)

=> " " " " " =6•9

=> x^3+y^3+z^3-3xyz=54