Using the formula for squaring a binomial, evaluate the following:

We shall use the identity (a-b)2 = a2 +b2 -2ab.

(i) $(69)^{2}$

$=(70-1)^{2}$

$=(70)^{2}-2 \times 70 \times 1+1$

$=4900-140+1$

$=4761$

(ii) $(78)^{2}$

$=(80-2)^{2}$

$=(80)^{2}-2 \times 80 \times 2+4$

$=6400-320+4$

$=6084$

(iii) $(197)^{2}$

$=(200-3)^{2}$

$=(200)^{2}-2 \times 200 \times 3+9$

$=40000-1200+9$

$=38809$

(iv) $(999)^{2}$

$=(1000-1)^{2}$

$=(1000)^{2}-2 \times 1000 \times 1+1$

$=1000000-2000+1$

$=998001$