# Evaluate the foll

(i) (99)3 = (100 – 1)3

Using identity,

(a – b)3 = a3 – b3 – 3ab(a – b)

(100 – 1)3 = (100)3 – 13 – (3 × 100 × 1) (100 – 1)

= 1000000 – 1 – 300 (100 – 1)

= 1000000 – 1 – 30000 + 300

= 970299

(ii) (102)3 = (100 + 2)3

Using identity,

(a + b)3 = a3 + b3 + 3ab (a + b)

(100 + 2)3 = (100)3 + 23 + (3 × 100 × 2) (100 + 2)

= 1000000 + 8 + 600 (100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)3 = (1000 – 2)3

Using identity,

(a – b)3 = a3 – b3 – 3ab (a – b)

(1000 – 2)3 = (1000)3 – 23 – (3 × 1000 × 2) (1000 – 2)

= 1000000000 – 8 – 6000 (1000 – 2)

= 1000000000 – 8 – 6000000 + 12000

= 994011992

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