Evaluate the following:
 $ 53\times 55 $

Hint: Use the identity $ (a+b)(a+c)={{a}^{2}}+(b+c)a+bc $ .
Choose the number $ a $ which is easier to multiply.
Observe that $ 53=50+3 $ and $ 55=50+5 $ , and multiplying by 50 is easier than finding squares of or multiplying by some other numbers (see Note below).

Complete step-by-step answer:
Since it is comparatively easier to multiply by 50, let us write the numbers as:
 $ 53\times 55=(50+3)(50+5) $
On multiplying by using the distributive property of multiplication, we get:
= $ {{(50)}^{2}}+(50)\times 5+3\times 5+3\times (50) $
On separating the common factors, we get:
= $ 2500+(3+5)\times 50+15 $
= $ 2500+8\times 50+15 $
= $ 2500+400+15 $
= $ 2915 $

Therefore, $ 53\times 55 $ is equal to 2915.

Note: We can also write $ 53\times 55=(54-1)(54+1)={{54}^{2}}-1 $. Calculating the square of 54 is now another task.
Also, $ 53\times 55=(55-2)55={{55}^{2}}-2\times 55 $ is quite easy if you use some identities.
For instance, $ {{55}^{2}}-2\times 55=3025-110=2915 $.
Some useful algebraic identities:
 $ {{(a-b)}^{2}}={{(b-a)}^{2}} $
 $ (a+b)(a-b)={{a}^{2}}-{{b}^{2}} $
 $ {{(a\pm b)}^{2}}={{a}^{2}}\pm 2ab+{{b}^{2}} $
 $ {{(a\pm b)}^{3}}={{a}^{3}}\pm 3ab(a\pm b)\pm {{b}^{3}} $
 $ (a\pm b)({{a}^{2}}\mp ab+{{b}^{2}})={{a}^{3}}\pm {{b}^{3}} $