{1/(2-√3)^2}+{1/(2+√3)^2}

So, basically by doing 2 different methods, I've also got 2 different answers.

[Method 1]

Substitute (2-√3) with variable a, and (2+√3) with variable b. By doing it, we get:

={1/(2-√3)^2}+{1/(2+√3)^2}
=(1/a^2)+(1/b^2)
=(b^2/a^2b^2)+(a^2/a^2b^2)
=(b^2+a^2)/a^2b^2
=(b^2+a^2)/(ab)^2

From here, we'll do the numerator b^2+a^2 first to minimize confusion:

=b^2+a^2
=b^2+a^2+2ba-2ba
=b^2+2ba+a^2-2ba
=(b+a)^2-2ba

Then, replace the variables a and b with their original numbers:

=(b+a)^2-2ba
={(2+√3)+(2-√3)}^2-2(2+√3)(2-√3)
=(2+√3+2-√3)^2-2(4-3)
=4^2-2
=16-2
=14

Therefore, the numerator is '14'.

Now, the denominator. Replace the variables a and b with their original numbers:

=(ab)^2
={(2-√3)(2+√3)}^2
=(4+2√3-2√3-3)^2
=1^2
=1

Therefore, the denominator is '1'.

Finally, the answer is:

=numerator/denominator
=14/1
=14

[Method 2]

={1/(2-√3)^2}+{1/(2+√3)^2}
={1/(4-4√3+3)}+{1/(4+4√3+3)}
={1/(7-4√3)}+{1/(7+4√3)}
={(7+4√3)-{(7-4√3)}/{(7-4√3)(7+4√3)}
={(7+4√3-7+4√3)}/{(7-4√3)(7+4√3)}
=8√3/(49+28√3-28√3-48)
=8√3/1
=8√3

Did I miscalculate something? If yes, where?