Question

If x+y+z=9 and xy+yz+zx=23 then find the value of [x 3 +y 3 +z 3 - 3xyz]?

Answer 1

x 3 + y 3 + z 3 - 3xyz = (x+y+z) 3 - 3 (x+y+z)(xy+yz+zx) = 9^3 - 3. 9. 23 = 729 - 621 = 108

Answer 2

Answer:

$Given:\ x+y+z=8 \\ xy+yz+zx=20 \\ To \: find\: x^{3}+y^{3}+z^{3}-3xyz =?$

Answer:

$x+y+z=8 --------eq 1 \\ xy+yz+zx= 20 ----eq2 \\ squaring\:both\: side \\ x^{2}+y^{2} +z^{2}+2xy+2yz +2zx \\ x^{2}+y^{2}+z^{2}+2(xy+yz +zx ) \\ x^{2}+y^{2}+z^{2}+2(20)=64 \\ x^{2}+y^{2}+z^{2}= 64-40 =24 \\by:\ eq2 \\ x^{3}+y^{3}+z^{3}-3xyz(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx) \\ x^{3}+y^{3}+ z^{3}-3xyz=8(24-20) \\ x^{3}+y^{3}+z^{3}+3xyx=8(4) \\ x^{3}+y^{3}+z^{3}-3xyz= 32 Answer.$