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(D) Let $x = 28$,$y = -15$,and $z = -13$.First,calculate the sum of $x, y,$ and $z$:$x + y + z = 28 + (-15) + (-13) = 28 - 28 = 0$.We know the algebra

Vedclass · Accepted Answer

$16380$

Vedclass · Answer

(D) Let $x = 28$,$y = -15$,and $z = -13$.First,calculate the sum of $x, y,$ and $z$:$x + y + z = 28 + (-15) + (-13) = 28 - 28 = 0$.We know the algebraic identity: If $x + y + z = 0$,then $x^{3} + y^{3} + z^{3} = 3xyz$.Substituting the values into the identity:$(28)^{3} + (-15)^{3} + (-13)^{3} = 3(28)(-15)(-13)$.Calculating the product:$3 	imes 28 = 84$$(-15) 	imes (-13) = 195$$84 	imes 195 = 16380$.Therefore,the value is $16380$.

Without actually calculating the cubes,find the value of each of the following: $(28)^{3} + (-15)^{3} + (-13)^{3}$

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