# How can I show that this complicated expression with square and cube roots reduces to the value 7?

How would I go about reducing the complicated-looking expression $$\sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}}$$ to show that it is equal to 7?

I came across the complicated expression while calculating the single real solution for the cubic equation $$z^3 - 33 z - 112 = 0$$

Using that cubic equation, I can show that $z = 7$ satisfies it and that a standard way of solving cubic equations shows that there is only a single real root and that its value is equal to the complicated expression, which is a very roundabout way of proving that the complicated expression is equal to 7. But what if I didn't know about the cubic equation? Is there a more direct way of reducing the complicated expression to a simpler one?

• Take the power 3 of this expression, using binomial formula $(a+b)^3$... – Jean Marie Feb 25 '17 at 21:56
• Note that if $p+q+r=0$ then $p^3+q^3+r^3=3pqr$ - you can use this to get rid of the cube roots by taking the two components as $p, q$ and the sum as $p+q=-r$. – Mark Bennet Feb 25 '17 at 22:16

I'll outline the basic method to simplify$$\sqrt[3]{56+19\sqrt{5}}+\dfrac {11}{\sqrt[3]{56+19\sqrt{5}}}\tag{1}$$ To simplify the nested radical, note that we have this general outline:$$\sqrt[n]{A+B\sqrt[m]{C}}=a+b\sqrt[m]{C}\tag{2}$$ So therefore, we have\begin{align*}\sqrt[3]{56+19\sqrt{5}} & =a+b\sqrt{5}\\56+19\sqrt{5} & =\underbrace{(a^3+15ab^2)}_{56}+\underbrace{(3a^2b+5b^3)}_{19}\sqrt{5}\end{align*} So we get this system of equations:\begin{align*} & a^3+15ab^2=56\\ & 3a^2b+5b^3=19\end{align*}\tag{3} $(3)$ has real solutions as $(a,b)=\left(\dfrac 72,\dfrac 12\right)$ so$$\sqrt[3]{56+19\sqrt{5}}=\dfrac {7+\sqrt5}2\tag4$$ Using $(4)$, and through some algebraic manipulations, we have\begin{align*}\color{blue}{\sqrt[3]{56+19\sqrt5}}+\dfrac {11}{\color{red}{\sqrt[3]{56+19\sqrt{5}}}} & =\color{blue}{\dfrac {7+\sqrt{5}}2}+\dfrac {11}{\color{red}{\frac {7+\sqrt{5}}2}}\\ & =\dfrac {7+\sqrt{5}}{2}+\dfrac {22}{7+\sqrt{5}}\\ & =\dfrac {7+\sqrt{5}}2+\dfrac {22(7-\sqrt5)}{44}\\ & =\dfrac {7+\sqrt5}2+\dfrac {7-\sqrt5}2\\ & =\boxed 7\end{align*} Just like what you got!

One possibility is to factor $19\sqrt{5}+56$ in the ring of integers of the field $\mathbb{Q}(\sqrt{5})$, which gives $$19\sqrt{5}+56 = \left(\frac{1+\sqrt{5}}2 \right)^3(4-\sqrt5)^3$$ and now it is easy to continue.

If you wonder how I arrived to this factorization, it is known that the ring of integers of $\mathbb Q(\sqrt{5})$ is a unique factorization domain, the norm of $19\sqrt{5}+56$ in this field is $11^3$ so this implies that it is divisible by either $4-\sqrt{5}$ or $4+\sqrt{5}$ which are the primes of $\mathbb Q(\sqrt{5})$ under $11$. After performing the division, the result is a unit, and so it is of the form $\pm (\tfrac{1+\sqrt{5}}{2})^k$, as $\tfrac{1+\sqrt{5}}{2}$ is the fundamental unit of $\mathbb Q(\sqrt{5})$.

Using $a=\sqrt[3]{19\sqrt5+56}$ and $b=\frac{11}{\sqrt[3]{19\sqrt5+56}}$ and $u=a+b$, we have \begin{align} a^3+b^3 &=19\sqrt5+56+\frac{1331}{19\sqrt5+56}\\ &=\frac{6272+2128\sqrt5}{56+19\sqrt5}\\[6pt] &=112 \end{align} Thus, \begin{align} &112\\ &=a^3+b^3\\ &=(a+b)\left(a^2-ab+b^2\right)\\ &=(a+b)\left((a+b)^2-3ab\right)\\ &=u(u^2-33) \end{align} Noting that for $x\lt\sqrt{11}$, \begin{align} x^3-33x-112 &\le22\sqrt{11}-112\\ &\lt0 \end{align} and for $x\ge\sqrt{11}$, $x^3-33x-112$ is increasing, we see that $x^3-33x-112=0$ has only one real solution, and that is $x=7$.

Therefore, $\sqrt[3]{19\sqrt5+56}+\frac{11}{\sqrt[3]{19\sqrt5+56}}=a+b=u=7$.

But what if I didn't know about the cubic equation?

As long as you have a reasonable suspicion that the expression might be rational, the polynomial it satisfies can be derived rather easily. Let:

$$x = \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}}$$

Then using $(a+b)^3=a^3+b^3+3ab(a+b)\,$ and $(56+19\sqrt{5})(56-19\sqrt{5})=1331=11^3\,$:

\require{cancel} \begin{align} x^3 & = 19\sqrt{5} + 56 + \frac{11^3}{19\sqrt{5} + 56} + 3 \cdot \cancel{\sqrt[3]{19\sqrt{5} + 56}} \cdot \frac{11}{\cancel{\sqrt[3]{19\sqrt{5} + 56}}} \cdot x \\ & = 56+ \cancel{19\sqrt{5}} + 56 - \cancel{19\sqrt{5}} + 33x \\ & = 112 + 33x \end{align}

By the rational root theorem, the equation $x^3-33x-112=0$ can only have rational roots that are integer divisors of $112=2^4\cdot 7\,$. Obviously $\pm 1$ do not satisfy the equation, and even roots can be eliminated by comparing the powers of $2$ between the terms, which leaves $\pm 7\,$ to try.

• Wish the downvoter had left a comment why. – dxiv Feb 26 '17 at 21:21