# Simplifying the expression $(1+\sqrt3)/(1-\sqrt3)+1/(1+\sqrt3)+2/(1+\sqrt{3})$

Can anyone give provide me some help to simplify this expression?

The three denominators are pretty much different, and I can't find a common denominator.

$$\frac{1+\sqrt3}{1-\sqrt3}+\frac1{1+\sqrt3}+\frac2{1+\sqrt{3}}$$

The calculator said it's equal to $-2$, but I don't get how a complicated expression like this would be equal to $-2$.

• You can use the brackets to make the fourth root, instead of 3^(1/4) write 3^{1/4}, alternatively, you can use \sqrt{3} to get $\sqrt{3}$. Jul 19, 2013 at 16:18
• Still not a polynomial.
– mrf
Jul 19, 2013 at 16:20
• The result is not $-2$, you sure you copied the question correctly? Here: wolframalpha.com/input/…
– user67258
Jul 19, 2013 at 16:21
• @DannyCheuk: your input has the last one with a fourth root instead of square root. Here is the one asked for. Still a ways from $-2$ Jul 19, 2013 at 16:53
• You're indeed correct, thanks for the correction!
– user67258
Jul 19, 2013 at 16:54

Here is a method.

This is not a polynomial. However ... the basic rules for combining fractions apply even in cases with roots in. You put the expression over a common denominator. Note that for any $x$ whatsoever $(1+x)(1-x)=1-x^2$.

You then have two fractions to add, and you will be able to use the same identity again.

Hint: $(1-a)(1+a)=1-a^2$ and hence $(1-a)(1+a)(1+a^2)=1-a^4$

Putting $\sqrt3=x$

$$\frac{1+\sqrt3}{1-\sqrt3}+\frac1{1+\sqrt3}+\frac2{1+\sqrt{3}}$$

$$=\frac{1+x}{1-x}+\frac1{1+x}+\frac2{1+x^2}$$

$$=\frac x{1-x}+\frac1{1-x}+\frac1{1+x}+\frac2{1+x^2}$$

$$=\frac x{1-x}+\frac{1+x+1-x}{(1-x)(1+x)}+\frac2{1+x^2}$$

$$=\frac x{1-x}+2\left(\frac1{1-x^2}+\frac1{1+x^2}\right)$$

$$=\frac x{1-x}+2\left(\frac{1+x^2+1-x^2}{(1-x^2)(1+x^2)}\right)$$

$$=\frac x{1-x}+\frac4{1-x^4}$$

Now $x^4=3$ as $\sqrt3=x$

• I'm having trouble going from the second line to the third. Jul 19, 2013 at 19:34
• Shouldn't the third line be $$\frac{(1+x)^2+1-x}{(1-x)(1+x)}+\frac2{1+x^2}$$ Jul 20, 2013 at 0:35
• @TheChaz2.0, sorry for the mistake.Please have a look into the edited version Jul 20, 2013 at 6:45
• @chubakueno, sorry for the mistake.Please have a look into the edited version.Thanks for your observation Jul 20, 2013 at 6:46

Multiply the first and second term by their conjugate respectively. Add the terms.

You will get

$\frac{2 + \sqrt{3}+\sqrt{3}}{1-\sqrt{3}}$ Now multiply both the bottom and top by the conjugate of the denominator. And do the same for the third term.