If x=z=225 and y=226,then find the value of x | vedclass

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(D) The algebraic identity for the expression is given by:$x^{3}+y^{3}+z^{3}-3xyz = \frac{1}{2}(x+y+z)[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}]$Given $x=225$,$y

Vedclass · Accepted Answer

$676$

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(D) The algebraic identity for the expression is given by:$x^{3}+y^{3}+z^{3}-3xyz = \frac{1}{2}(x+y+z)[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}]$Given $x=225$,$y=226$,and $z=225$:Substitute these values into the identity:$= \frac{1}{2}(225+226+225)[(225-226)^{2}+(226-225)^{2}+(225-225)^{2}]$$= \frac{1}{2}(676)[(-1)^{2}+(1)^{2}+(0)^{2}]$$= \frac{1}{2} 	imes 676 	imes [1 + 1 + 0]$$= \frac{1}{2} 	imes 676 	imes 2$$= 676$

If $x=z=225$ and $y=226$,then find the value of $x^{3}+y^{3}+z^{3}-3xyz$.

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