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

Solution:

(i) Here, we will use the identity $(a+b)^{2}=a^{2}+2 a b+b^{2}$

$(102)^{2}$

$=(100+2)^{2}$

$=(100)^{2}+2 \times 100 \times 2+2^{2}$

$=10000+400+4$

$=10404$

(ii) Here, we will use the identity $(a-b)^{2}=a^{2}-2 a b+b^{2}$

$(99)^{2}$

$=(100-1)^{2}$

$=(100)^{2}-2 \times 100 \times 1+1^{2}$

$=10000-200+1$

$=9801$

(iii) Here, we will use the identity $(a+b)^{2}=a^{2}+2 a b+b^{2}$

(1001) $^{2}$

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

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

$=1000000+2000+1$

$=1002001$

(iv) Here, we will use the identity $(a-b)^{2}=a^{2}-2 a b+b^{2}$

$(999)^{2}$

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

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

$=1000000-2000+1$

$=998001$

(v) Here, we will use the identity $(a+b)^{2}=a^{2}+2 a b+b^{2}$

(703) $^{2}$

$=(700+3)^{2}$

$=(700)^{2}+2 \times 700 \times 3+3^{2}$

$=490000+4200+9$

$=494209$