We have to simplify [(a+i)^3-(a-i)^3]/[(a+i)^2-(a-i)^2].

[(a+i)^3-(a-i)^3]/[(a+i)^2-(a-i)^2]

we open the brackets

=> [a^3 + 3a^2*i+ 3a*i^2+ i^3 - a^3 + 3*a^2*i - 3*a*i^2 + i^3]/[(a^2 + 2ai + i^2 - a^2 - i^2 + 2ai]

cancel the common terms in the numerator and denominator

=> [ 3a^2*i+ i^3 + 3*a^2*i + i^3]/[2ai + 2ai]

=> [6a^2*i+ 2i^3 ]/[4ai]

=> [3a^2*i+ i^3 ]/[2ai]

=> [3a^2*i - i ]/[2ai]

=> [3a^2 - 1 ]/[2a]

=> ( 3a^2 - 1)/ 2a

**The required simplified form is ( 3a^2 - 1)/ 2a**

We'll re-write the difference of cubes from numerator:

(a+i)^3-(a-i)^3 = (a + i - a + i)[(a+i)^2 + (a+i)(a-i) + (a-i)^2]

We'll combine and eliminate like terms:

(a+i)^3-(a-i)^3 = (2 i)(a^2 + 2ai + i^2 + a^2 - i^2 + a^2 - 2ai + i^2)

(a+i)^3-(a-i)^3 = (2 i)(3a^2 + i^2)

(a+i)^3-(a-i)^3 = (2 i)(3a^2 -1)

We'll re-write the difference of squares from numerator:

[(a+i)^2-(a-i)^2] = (a + i - a + i)(a + i + a - i)

We'll combine and eliminate like terms:

[(a+i)^2-(a-i)^2] = 2i*2a

We'll re-write the fraction:

(2 i)(3a^2 -1)/2i*2a = (3a^2 -1)/2a

3a^2 -1 = (a*sqrt3 - 1)(a*sqrt3 + 1)

**[(a+i)^3-(a-i)^3]/[(a+i)^2-(a-i)^2] = [(a*sqrt3 - 1)(a*sqrt3 + 1)]/2a**