Find the value of x^{3}+y^{3}+z^{3}-3xyz when | vedclass

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(C) 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 val

Vedclass · Accepted Answer

$4940$

Vedclass · Answer

(C) 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 values are $x=89, y=87, z=84$.Step $1$: Calculate the sum $(x+y+z)$:$x+y+z = 89+87+84 = 260$Step $2$: Calculate the differences:$(x-y) = 89-87 = 2$$(y-z) = 87-84 = 3$$(z-x) = 84-89 = -5$Step $3$: Substitute these values into the identity:$= \frac{1}{2} 	imes 260 	imes \{2^{2} + 3^{2} + (-5)^{2}\}$$= 130 	imes \{4 + 9 + 25\}$$= 130 	imes 38$$= 4940$

Find the value of $x^{3}+y^{3}+z^{3}-3xyz$ when $x=89, y=87, z=84$.

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