If x=332, y=333, z=335, then the value of x^{ | vedclass

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

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$7000$

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(A) The algebraic identity for $x^{3}+y^{3}+z^{3}-3xyz$ is given by:$x^{3}+y^{3}+z^{3}-3xyz = \frac{1}{2}(x+y+z)[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}]$Given $x=332, y=333, z=335$.First,calculate the sum: $x+y+z = 332+333+335 = 1000$.Next,calculate the differences:$(x-y) = 332-333 = -1 \implies (x-y)^{2} = 1$$(y-z) = 333-335 = -2 \implies (y-z)^{2} = 4$$(z-x) = 335-332 = 3 \implies (z-x)^{2} = 9$Now,substitute these values into the identity:$= \frac{1}{2} 	imes 1000 	imes (1 + 4 + 9)$$= 500 	imes 14$$= 7000$

If $x=332, y=333, z=335,$ then the value of $x^{3}+y^{3}+z^{3}-3xyz$ is

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