Determining a limit of a function doesnt match up

Question

Consider this function

$$\lim_{x \to 2} \frac {x^3 - 8} {x-2}$$

Now what I've done is I've tried to alter the fraction to avoid division by zero. For the numerator, I've put it as this

$$ \frac {(x - 2)^3} {x-2} $$

After this I've simply gotten rid of the cube and the lower part of the fraction leaving me with this

$$(x-2)^2$$

And here I've applied the formula $$a^2 -2ab + b^2$$ And I than I put in 2 and I get as a result $$ x^2 +4x + 4$$ If i put in 2 now I get 16. Now the solutions say 12, but the solutions are notorius to be wrong more often than you would think.Is my solution correct? If not what am I doing wrong.

Thanks!

Answer 1

$x^3-a^3\ne (x-a)^3$, except in characteristic $3$.

It can be much simpler to determine: $\frac{x^3-2^3}{x-2}$ is the rate of variation of the function $x^3$ from $x=2$, so the limit is its derivative $\:3x^2\bigr|_{x=2}=12$.