= lim (x-->infinity) [(x^2 - 1) - (x^3 - 1)^(2/3)]/[√(x^2 - 1) + (x^3 - 1)^(1/3)]
= lim (x-->infinity) [(x^2 - 1)^3 - (x^3 - 1)^2]/{[√(x^2 - 1) + (x^3 - 1)^(1/3)][(x^2 - 1)^2 + (x^2 - 1)(x^3 - 1)^(2/3) + (x^3 - 1)^(4/3)]}
= lim (x-->infinity) (-3x^4 + 2x^3 + 3x^2 - 2)/{[√(x^2 - 1) + (x^3 - 1)^(1/3)][(x^2 - 1)^2 + (x^2 - 1)(x^3 - 1)^(2/3) + (x^3 - 1)^(4/3)]}.
Now, at this point, we can use the fact that for x ≠ 0:
(a) √(x^2 - 1) - (x^3 - 1)^(1/3) = √[x^2(1 - 1/x^2)] - [x^3(1 - 1/x^3)]^(1/3)
= √(x^2)√(1 - 1/x^2) - (x^3)^(1/3)(1 - 1/x^3)^(1/3)
= x√(1 - 1/x^2) - x(1 - 1/x^3)^(1/3).
(b) (x^2 - 1)^2 - (x^2 - 1)(x^3 - 1)^(2/3) + (x^3 - 1)^(4/3)
= [x^2(1 - 1/x^2)]^2 - [x^2(1 - 1/x^2)][x^3(1 - 1/x^3)]^(2/3) + [x^3(1 - 1/x^3)]^(4/3)
= x^4(1 - 1/x^2)^2 - x^4(1 - 1/x^3) + x^4(1 - 1/x^3)^(4/3).
Putting this altogether gives:
lim (x-->infinity) (-3x^4 + 2x^3 + 3x^2 - 2)/{[√(x^2 - 1) + (x^3 - 1)^(1/3)][(x^2 - 1)^2 + (x^2 - 1)(x^3 - 1)^(2/3) + (x^3 - 1)^(4/3)]}
= lim (x-->infinity) (-3x^4 + 2x^3 + 3x^2 - 2)/{[x√(1 - 1/x^2) + x(1 - 1/x^3)^(1/3)][x^4(1 - 1/x^2)^2 - x^4(1 - 1/x^3) + x^4(1 - 1/x^3)^(4/3)]}
= lim (x-->infinity) (-3 + 2/x + 3/x^2 - 2/x^4)/{[x√(1 - 1/x^2) + x(1 - 1/x^3)^(1/3)][(1 - 1/x^2)^2 - (1 - 1/x^3) + (1 - 1/x^3)^(4/3)]}, by dividing the numerator and denominator by x^4.
The numerator goes to -3, while the denominator goes to infinity. Therefore, this limit is 0.
(4) can be done by combining fractions (the LCD is x - 1).
I hope this helps!