Question

if x+y+z=9 and xy+yz+zx=23.find the valure of x3 +y3 + z3-3xyz

Answer 1

in these question first find the value of a^2+b^2+c^3
with the help of the( a+b+c)^2
then put it on that formula (a^3+b^3+c^3-3abc) u will get ur ans

Answer 2

Answer:

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

Answer:

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

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