Simplifying Radicals

**rtblue** · November 25th 2010, 08:13 AM

The problem is:

$\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$

I have tried to factor it, but that doesn't seem to work. Multiplying by the conjugate looks like it involves much more work than is necessary. Any suggestions? Help is appreciated.

**Opalg** · November 26th 2010, 12:01 AM

Originally Posted by rtblue

The problem is:

$\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$

I have tried to factor it, but that doesn't seem to work. Multiplying by the conjugate looks like it involves much more work than is necessary. Any suggestions? Help is appreciated.

My first reaction was to evaluate this expression using a calculator, to see if that gave any clues. The answer came out to be exactly(?) 1. So if the answer really is as tidy as that, there should be some way to find it without too much work.

Let $x = \sqrt[3]{2+\sqrt{5}}$ and $y = \sqrt[3]{2-\sqrt{5}}$ . Then $x^3+y^3 = 4$ , which factorises as $(x+y)(x^2-xy+y^2) = 4$ . We want to find $x+y$ , so let $x+y=z$ . Then

$z(x^2-xy+y^2) = 4.\quad(*)$

Also, $xy = \sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})} = \sqrt[3]{-1} = -1$ . Hence $z^2 = (x+y)^2 = x^2+2xy+y^2 = x^2+y^2-2$ , so that $x^2+y^2 = z^2+2$ , and equation (*) becomes $z(z^2+3) = 4$ , or $z^3+3z=4$ . The only real solution to that cubic equation is z=1, which is what we wanted.

Taking this a bit further, the fact that x and y satisfy the equations $xy=-1$ and $x+y=1$ imply that $x^2-x-1=0$ , which is the equation for the golden ratio $\frac12(1+\sqrt5)$ . Thus $x = \frac12(1+\sqrt5)$ , as you can verify by checking that $\bigl(\frac12(1+\sqrt5)\bigr)^3 = 2+\sqrt5$ .

**HallsofIvy** · November 26th 2010, 02:56 AM

That particular form reminds me of Cardano's "cubic formula".

$(a+ b)^3= a^2+ 3a^2b+ 3ab^2+ a^3$
$3ab(a+ b)= 3a^2b+ 3ab^2$
so $(a+ b)^3- 3ab(a+ b)= a^3- b^3$

If you let x= a+ b, m= 3ab and $n= a^3- b^3$ then $x^3- mx= n$ , a "reduced" cubic.

Suppose we know m and n. Can we solve for a and b and so find x? Yes, we can!

From m= 3ab, $b= \frac{m}{3a}$ so $a^3- b^3= a^3- \frac{m^3}{3^3a^3}= n$ . Multiplying through by $a^3$ gives the quadratic equation, in $a^3$ , $(a^3)^2- \frac{m^3}{3^3}= na^3$ or $(a^3)^2- na^3- \frac{m^3}{3^3}= 0$ . By the quadratic formula, $a^3= \frac{n\pm\sqrt{n^2+ 4\frac{m^3}{3^3}}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$ .

From $a^3- b^3= n$ , $b^3= a^3- n= -\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$ .

$\sqrt[3]{2+ \sqrt{5}}$ looks just like "a", above, with $\frac{n}{2}= 2$ so $n= 4$ and $\sqr{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}= \sqrt{\left(\frac{4}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$ $= \sqrt{4+ \left(\frac{m}{3}\right)^3}= \sqrt{5}$ so that $\frac{m}{3}= 1$ and m= 3.

That is, $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ is a root of $x^3+ 3x= 4$ or $x^3+ 3x- 4= 0$ . It is easy to see, say by graphing, that that equation has only one real root and, of course, $1^3+ 3(1)- 4= 0$ so that single real root, and so the real value of the original expression, is 1.

**Soroban** · November 26th 2010, 03:10 AM

Hello, rtblue!

$\text{The problem is: }\:\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$

Whenever I see the form $(a \pm b\sqrt{5})$ ,
. . I suspect that the Golden Mean is embedded somewhere.

I know from experience that $(3 + \sqrt{5})$ arises from $\phi^2$ .

I just learned that:
. . $\displaystyle \phi^3 \:=\:\left(\frac{1 + \sqrt{5}}{2}\right)^3 \:=\:\frac{1 + 3\sqrt{5} + 15 + 5\sqrt{5}}{8} \:=\:\frac{16 + 8\sqrt{5}}{8} \:=\:2 + \sqrt{5}$

The problem becomes:

. . $\displaystyle \sqrt[3]{\left(\frac{1 + \sqrt{5}}{2}\right)^3} + \sqrt[3]{\left(\frac{1-\sqrt{5}}{2}\right)^3} \;=\;\frac{1+\sqrt{5}}{2} + \frac{1-\sqrt{5}}{2} \;=\;1$

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