Calculus Limits

: Limit[(x^3 - a^3 / (x^4 - a^4), x->a]
Use difference of two cubes to factor numerator
: Factor[(x^3 - a^3)]
: - (a - x) (a^2 + a * x + x^2)
Use substitution and difference of two squares to factor denominator (x^2)^2 - (z^2)^2
:Factor[(x^4 - a^4)]
:- (a - x) * (a + x) * (a^2 + x^2)
Cancel terms
:( - (a - x) * (a^2 + a * x + x^2)) / (- (a - x) (a + x) * (a^2 + x^2))
:(a^2 + a * x + x^2) / ((a + x) * (a^2 + x^2))
The Limit of a product is is product of the limits
:Limit[(a^2 + a * x + x^2) / ((a + x) * (a^2 + x^2)), x
:-> a] == Limit[a^2 + a * x + x^2, x -> a] * Limit[1 / ((a + x) *(a^2 :+ x^2)), x -> a]
Numerator
:a^2 + a * x + x^2 /. x -> a
:3 * a^2
:Limit[a^2 + a * x + x^2, x -> a]
:Num := 3 * a^2
Denominator
:(a + x) *(a^2 + x^2) /. x -> a
:4 * a^3
:Limit[1 / ((a + x) * (a^2 + x^2)), x -> a]
:Den := 1 / (4 * a^3)
:Limit[(a^2 + a * x + x^2) / ((a + x) * (a^2 + x^2)), x -> a] ==
:Num * Den
:True
:Limit[(a^2 + a * x + x^2) / ((a + x) * (a^2 + x^2)), x -> a]
Answer
:3 / (4 * a)

The answer provided a couple of days ago is quite correct. I just wanted to add that the problem appears to be leading you to recognize the common factor in "telescoping" functions. In general the telescoping function looks like this:
a^n-b^n=(a-b)(a^(n-1)+a^(n-2)*b+ ... + a*b^(n-2)+ b^(n-1))
When you expand the right hand side you see that a*a^(n-1) is a^n and that -b*b^(n-1) is -b^n and that all the other terms cancel out. for instance -b*a^(n-1)+a*a^(n-2)*b = 0
In your case you have two examples of the telescoping function dividing each other: (x^3-a^3)/(x^4-a^4). From the telescoping function these both have (x-a) as a common factor and you can cancel it out to help compute the limit. Note that the x=a must be still be excluded from the domain but when taking limits we are merely approaching the value.

(x^3-a^3)/(x^4-a^4) = ( (x-a)(x^2 + xa + a^2) )/( (x-a)(x^3 + x^2*a + x*a^2 + a^3) )
we cancel out the (x-a) and get this
= (x^2 + xa + a^2)/(x^3 + x^2*a + x*a^2 + a^3)
then let x -> a
we get (a^2 + a^2 + a^2)/(a^3 + a^3 + a^3+ a^3)
= 3*a^2/(4a^3) = 3/(4a)

Which is the same answer provided earlier but with a slightly different approach

If two roots of x cube + 3 x square minus minus plus size zero are equal then find the value of c