# 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 $-2$.

• You can use the brackets to make the fourth root, instead of 3^(1/4) write 3^{1/4}, alternatively, you can use \sqrt[4]{3} to get $\sqrt[4]{3}$. Commented Jul 19, 2013 at 16:18
• Still not a polynomial.
– mrf
Commented Jul 19, 2013 at 16:20
• The result is not $-2$, you sure you copied the question correctly? Here: wolframalpha.com/input/…
– user67258
Commented 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 $-2$ Commented Jul 19, 2013 at 16:53
• You're indeed correct, thanks for the correction!
– user67258
Commented 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 $(1+x)(1-x)=1-x^2$.

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.

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

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 $\sqrt[4]3=x$

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

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

You will get

$\frac{2 + \sqrt[4]{3}+\sqrt{3}}{1-\sqrt{3}}$ Now multiply both the bottom and top by the conjugate of the denominator. And do the same for the third term.