Evaluate the following using suitable identities:

 (i) (99)3

(ii) (102)3

(iii) (998)3

Solutions:

(i) (99)3

Solution:

We can write 99 as 100–1

Using identity, (x –y)3 = x3–y3–3xy(x–y)

(99)3 = (100–1)3

= (100)3–13–(3×100×1)(100–1)

= 1000000 –1–300(100 – 1)

= 1000000–1–30000+300

= 970299

(ii) (102)3

Solution:

We can write 102 as 100+2

Using identity,(x+y)3 = x3+y3+3xy(x+y)

(100+2)3 =(100)3+23+(3×100×2)(100+2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)3

Solution:

We can write 99 as 1000–2

Using identity,(x–y)3 = x3–y3–3xy(x–y)

(998)3 =(1000–2)3

=(1000)3–23–(3×1000×2)(1000–2)

= 1000000000–8–6000(1000– 2)

= 1000000000–8- 6000000+12000

= 994011992