Question

(99)^3 evaluate using suitable identities
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Answer 1

(99)

3

=(100−1)

3

We know that

(a−b)

3

=a

3

−b

3

−3ab(a−b)

Therefore,

=(100)

3

−1

3

−3×100×1×(100−1)

=1000000−1−300(100−1)

=1000000−1−30000+300

=970299

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Answer 2

Answer:

(a-b)³= a³-b³-3ab(a-b)

(99)³=(100-1)³

(99)³=1000000-1-300(100-1)

(99)³=1000000-1-29700

(99)³=970299

PLS MARK BRAINLIEST