Simplifying the expression $(1+\sqrt[4]3)/(1-\sqrt[4]3)+1/(1+\sqrt[4]3)+2/(1+\sqrt{3})$ 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.
$$\frac{1+\sqrt[4]3}{1-\sqrt[4]3}+\frac1{1+\sqrt[4]3}+\frac2{1+\sqrt{3}}$$
The calculator said it's equal to $-2$, but I don't get how a complicated expression like this would be equal to $-2$.
 A: 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 $(1+x)(1-x)=1-x^2$.
Use this to help you to add the first pair of fractions and simplify it.
You then have two fractions to add, and you will be able to use the same identity again.
A: Hint: $(1-a)(1+a)=1-a^2$ and hence $(1-a)(1+a)(1+a^2)=1-a^4$
A: Putting  $\sqrt[4]3=x$
$$\frac{1+\sqrt[4]3}{1-\sqrt[4]3}+\frac1{1+\sqrt[4]3}+\frac2{1+\sqrt{3}}$$
$$=\frac{1+x}{1-x}+\frac1{1+x}+\frac2{1+x^2}$$
$$=\frac x{1-x}+\frac1{1-x}+\frac1{1+x}+\frac2{1+x^2}$$
$$=\frac x{1-x}+\frac{1+x+1-x}{(1-x)(1+x)}+\frac2{1+x^2}$$
$$=\frac x{1-x}+2\left(\frac1{1-x^2}+\frac1{1+x^2}\right)$$
$$=\frac x{1-x}+2\left(\frac{1+x^2+1-x^2}{(1-x^2)(1+x^2)}\right)$$
$$=\frac x{1-x}+\frac4{1-x^4}$$
Now $x^4=3$ as  $\sqrt[4]3=x$
A: Multiply the first and second term by their conjugate respectively. Add the terms.
You will get
$\frac{2 + \sqrt[4]{3}+\sqrt{3}}{1-\sqrt{3}}$
Now multiply both the bottom and top by the conjugate of the denominator. And do the same for the third term.
