How can I prove that {1÷ [2+(3) ^0.5]} is equal to [2-(3) ^0.5] by algebraic rules?

Answer 1

Recall: The difference of two squares is: A² - B² = (A - B)*(A + B)

In this problem, A = 2 and B = sqrt(3)

So: 2² - (sqrt(3))² = 4 - 3 = 1

To use these facts, just multiply both sides by this:

1 / (2 - sqrt(3))

Answer 2

Use the conjugate on the left hand side.

1/(2+√3) so multiply top and bottom by (2-√3)

Effectively, in the denominator you get a difference of squares which yield 4 - (√3)². This equals one and the top (numerator) is 2-√3. And, this matches the right hand of the equation.