Evaluate using su

(i) 53 × 55

We can re–write 53 and 55 as

(50+3)× (50+5)

Using the identity

(x+a) (x+b) = x2 + (a+b) x + ab

(50+3)× (50+5) where x = 50, a = 3 and b = 5

= 502 + ( 3+5) 50 + 3× 5

=2500 + 400 + 15

= 2915

(ii) 102 × 106

= (100 + 2) (100 + 6)

Using the identity

(x+a) (x+b) = x2 + (a+b) x + ab

Here x= 100, a = 2 and b = 6

(100 + 2) (100 + 6)

= 1002 + ( 2+6) 100 + 2× 6

= 10000 + 800 +12

= 10812

(iii) 34 × 36

= (30+4) + (30+6)

Using the identity

(x+a) (x+b) = x2 + (a+b) x + ab

Here x = 30, a= 4 and b = 6

So, 302 +( 4+6) 30 +(4× 6)

= 900 + 300 +24

= 1224

(iv) 103 × 96

= (90 + 13) (90 +6)

Using the identity

(x+a) (x+b) = x2 + (a+b) x + ab

Here x = 90, a = 13 and b = 6

So, (90 + 13) (90 +6)

= 902 + ( 13+6) 90 +(13× 6)

= 8100 + 1710 +78

= 9888

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