Question

If x + y+z=9 &xy+yz+zx=23 then the value of x^3+y^3+z^3-3xyz=?

Answer 1

Given x + y + z = 9 , xy + yz + zx = 23

 Consider, x + y + z = 9

 Squaring on both sides,

we get (x + y + z)2 = 81

x2 + y2 + z2 +2(xy + yz +zx) = 81

x2 + y2 + z2 = 81 − 2(xy + yz +zx) = 81−

2(23) = 81 − 46 = 35 

∴ x2 + y2 + z2 = 35

 We know that

x3 + y3 + z3 − 3xyz = (x + y + z)( x2 + y2 + z2 − xy − yz − zx) = 9(35 −23) = 9(12) = 32

x2 and others are read as x square

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Answer 2

Answer:

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

Answer:

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

Step-by-step explanation:

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