Question

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

Answer 1

\left[x _{1}\right] = \left[ \frac{\sqrt[3]{\left( \frac{\left( -27\right) \,y^{3}}{2}+\frac{\left( -27\right) \,z^{3}}{2}+\frac{\sqrt{\left( 729\,y^{6} - 2916\,yz^{3}+1458\,y^{3}\,z^{3}+729\,z^{6}\right) }}{2}\right) }}{3}+\frac{3\,\sqrt[3]{2}\,yz}{\sqrt[3]{\left( \left( -27\right) \,y^{3} - 27\,z^{3}+\sqrt{\left( 729\,y^{6} - 2916\,yz^{3}+1458\,y^{3}\,z^{3}+729\,z^{6}\right) }\right) }}\right][x​1​​]=​⎣​⎢​⎢​⎢​⎢​⎡​​​3​​​3​​√​(​2​​(−27)y​3​​​​+​2​​(−27)z​3​​​​+​2​​√​(729y​6​​−2916yz​3​​+1458y​3​​z​3​​+729z​6​​)​​​​​)​​​​​+​​3​​√​((−27)y​3​​−27z​3​​+√​(729y​6​​−2916yz​3​​+1458y​3​​z​3​​+729z​6​​)​​​)​​​​​3​3​​√​2​​​yz​​​⎦​⎥​⎥​⎥​⎥​⎤​​ Algebra functional value

Answer 2

Answer:

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

Answer:

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