Evaluate each of the following using identities:

Solution:

(i) We have,

$399^{2}=(400-1)^{2}$

$=(400)^{2}+(1)^{2}-2 \times 400 \times 1 \quad\left[(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

Where, a = 400 and b = 1

= 160000 + 1 - 8000

= 159201

Therefore, $(399)^{2}=159201$.

(ii)  We have,

$(0.98)^{2}=(1-0.02)^{2}$

$=(1)^{2}+(0.02)^{2}-2 \times 1 \times 0.02$

= 1 + 0.0004 - 0.04   [Where, a = 1 and b = 0.02]

= 1.0004 - 0.04

= 0.9604

Therefore, $(0.98)^{2}=0.9604$

(iii) 991 × 1009

Solution:

We have,

991 × 1009

= (1000 - 9)(1000 + 9)

$=(1000)^{2}-(9)^{2} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$

= 1000000 - 81        [Where a = 1000 and b = 9]

= 999919

Therefore, 991 × 1009 = 999919

(iv)  We have,

117 × 83

= (100 + 17)(100 - 17)

$=(100)^{2}-(17)^{2} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$

= 10000 - 289        [Where a = 100 and b = 17]

= 9711

Therefore, 117 × 83 = 9711