(D) We have $99 = 100 - 1$.Using the algebraic identity $(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b)$,where $a = 100$ and $b = 1$:$(100 - 1)^{3} = 100^{3

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(D) We have $99 = 100 - 1$.Using the algebraic identity $(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b)$,where $a = 100$ and $b = 1$:$(100 - 1)^{3} = 100^{3} - 1^{3} - 3(100)(1)(100 - 1)$$= 1000000 - 1 - 300(99)$$= 1000000 - 1 - 29700$$= 1000000 - 29701$$= 970299$