Question

O Evaluate using suitable identity
99³

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Answer 1

Answer:

We have,

(99)3

=(100−1)3

 

We know that

(a−b)3=a3−b3−3ab(a−b)

 

Therefore,

=(100)3−13−3×100×1×(100−1)

=1000000−1−300(100−1)

=1000000−1−30000+300

=970299

Answer 2

Answer:

99³= ( 100-1)³

(a-b) ³=a³-b³-3ab(a-b)

=100³-1³-3*100*1(100-1)

=1000000-1-300(100-1)

=1000000-301(100-1)

=1000000-30100+301

=1000000+301-30100

=1000301-30100

=970299

Explanation:

Hope it helps