If x+y+z=9 and xy+yz+zx=23,then the value of | vedclass

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(A) The algebraic identity for $x^3+y^3+z^3-3xyz$ is given by:$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$We know that $(x+y+z)^2 = x^2+y^2+z^2+

Vedclass · Accepted Answer

$108$

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(A) The algebraic identity for $x^3+y^3+z^3-3xyz$ is given by:$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$We know that $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$.Therefore,$x^2+y^2+z^2 = (x+y+z)^2 - 2(xy+yz+zx)$.Substituting this into the identity:$x^3+y^3+z^3-3xyz = (x+y+z)[(x+y+z)^2 - 2(xy+yz+zx) - (xy+yz+zx)]$$x^3+y^3+z^3-3xyz = (x+y+z)[(x+y+z)^2 - 3(xy+yz+zx)]$Given $x+y+z=9$ and $xy+yz+zx=23$:$= 9 	imes [9^2 - 3(23)]$$= 9 	imes [81 - 69]$$= 9 	imes 12 = 108$

If $x+y+z=9$ and $xy+yz+zx=23$,then the value of $x^3+y^3+z^3-3xyz$ is

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