# Create a monic polynomial with integer coefficients with $\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}$ as a root.

Create a monic polynomial with integer coefficients with $\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}$ as a root.

I want to create such polynomial but all the ways I have tried don't give a polynomial with integer coefficients and all of them give me irrational or non integer rationals. How should I work?

Please use high school math for solution and don't use matrix.

• How do you propose to do that Taha--i.e. do this without the standard methods? Can you provide context as to how you think this should go or why you think it is possible? Dec 8, 2016 at 18:29
• Is Lagrange interpolation acceptable? Dec 8, 2016 at 18:30
• @EliSadoff Yes of course But I have tried it it will gave a non integer polynomial. Dec 8, 2016 at 18:31
• I don't really know of any methods that would work that both give you an integer polynomial and do not use a matrix. Try Chebyshev maybe? Dec 8, 2016 at 18:31
• A link to math.stackexchange.com/questions/2041887/… might be in order. There is an answer there that gives the polynomial, without deriving it. I assume you'd like to see a simple, step-by-step procedure for the derivation. Dec 8, 2016 at 18:36

The systematic approach would be to write $x=\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}$ as a system of algebraic equations:

\begin{align} \begin{cases} x - a - b - c & = 0 \\ a^3 - 2 & = 0 \\ b^3 - 3 & = 0 \\ c^3 - 5 & = 0 \end{cases} \end{align}

Eliminating $a,b,c$ between the equations will give a single equation in $x$ with integer coefficients.

Elimination can be done by repeatedly using the resultant of two polynomials, which is a routine calculation, thogh not pretty to do by hand in this particular case. Wolfram Alpha resultant[ resultant[ resultant[ x - a - b - c, a^3 - 2, a], b^3 - 3, b ], c^3 - 5, c] gives:

$$-x^{27} + 90 x^{24} - 1089 x^{21} + 62130 x^{18} - 105507 x^{15} + 16537410 x^{12} + 30081453 x^{9} + 1886601330 x^{6} - 73062900 x^{3} + 6859000$$

Write $a^3=2, b^3=3, c^3=5$. Whenever we get powers $3$ or more for one of these, use these to reduce to power $2$ or less.

Begin: $$x = a+b+c$$ Step 1
Solve for $a$, then cube, multiply out, then reduce $$a=b+c-x \\ a^3=(b+c-x)^3 \\ 2=-3\,{b}^{2}c+3\,{b}^{2}x-3\,b{c}^{2}+6\,bcx-3\,b{x}^{2}+3\,{c}^{2}x- 3\,c{x}^{2}+{x}^{3}-8$$ We have eliminated $a$

Step 2
Now solve for $b$. Oops it is quadratic in $b$. So anyway $$\left( -3\,c+3\,x \right) {b}^{2}+ \left( -3\,{c}^{2}+6\,cx-3\,{x}^{2 } \right) b+3\,{c}^{2}x-3\,c{x}^{2}+{x}^{3}-10=0$$ Multiply by $b$ and reduce to get another quadratic in $b$ $$\left( -3\,{c}^{2}+6\,cx-3\,{x}^{2} \right) {b}^{2}+ \left( 3\,{c}^{2 }x-3\,c{x}^{2}+{x}^{3}-10 \right) b-9\,c+9\,x =0$$ Now eliminate the $b^2$ term: multiply the first quadratic by $(x-c)$ and add to the second quadratic, so that the $b^2$ terms cancel. Reduce and solve for $b$: $$b ={\frac {6\,{c}^{2}{x}^{2}-4\,c{x}^{3}+{x}^{4}+c-16\,x}{6\,{c}^{2}x-6\, c{x}^{2}+2\,{x}^{3}-5}}$$ Now cube this to eliminate $b$.

Step 3
Now do it all again for $c$. We have $$\left( 66\,{x}^{10}-4392\,{x}^{7}+17955\,{x}^{4}-1398\,x \right) {c}^ {2}+ \left( -12\,{x}^{11}+2583\,{x}^{8}-26568\,{x}^{5}+10308\,{x}^{2} \right) c+{x}^{12}-1112\,{x}^{9}+28008\,{x}^{6}-39886\,{x}^{3}+380=0$$ Multiply by $c$ and reduce to get another quadratic in $c$ $$\left( -12\,{x}^{11}+2583\,{x}^{8}-26568\,{x}^{5}+10308\,{x}^{2} \right) {c}^{2}+ \left( {x}^{12}-1112\,{x}^{9}+28008\,{x}^{6}-39886\, {x}^{3}+380 \right) c+330\,{x}^{10}-21960\,{x}^{7}+89775\,{x}^{4}-6990 \,x=0$$ Now eliminate the $c^2$ term: multiply the first quadratic by $(4x^9-861x^6+8856x^3-3436)x$ and the second by $22x^9-1464x^6+5985x^3-466$ then add so that the $c^2$ terms cancel. $$\left( -26\,{x}^{21}-5264\,{x}^{18}-186378\,{x}^{15}-2704430\,{x}^{12 }-26489048\,{x}^{9}-69750462\,{x}^{6}-14557112\,{x}^{3}-177080 \right) c+4\,{x}^{22}+1951\,{x}^{19}+112080\,{x}^{16}+1973800\,{x}^{ 13}+23034286\,{x}^{10}+107977011\,{x}^{7}+56743276\,{x}^{4}+1951660\,x =0$$ Finally, solve for $c$, then cube to eliminate $c$. We end up with an equation in $x$ and integers only.

(Posting this as a separate answer, since unlike my previous one which was more generic and higher level, the following is narrower and fully hands-on.)

Let:

\begin{align} x & = \sqrt[3]{2} + \sqrt[3]{3} + \sqrt[3]{5} \tag{1}\\ y & = \sqrt[3]{2\cdot 3} + \sqrt[3]{3\cdot 5} + \sqrt[3]{5\cdot 2} \tag{2} \\ z & = \sqrt[3]{2\cdot 3\cdot 5} \quad\iff\quad z^3 = 30\tag{3} \end{align}

The following identity can be derived from Newton's relations, or otherwise be verified directly:

$$a^3+b^3+c^3 = (a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc \tag{4}$$

Using the identity above with the terms of $x,y$ gives:

\begin{align} 10 = 2+3+5 & = x^3-3 x y + 3z \tag{5}\\ 31 = 2\cdot 3+3\cdot 5+ 5 \cdot 2 &= y^3-3y(\sqrt[3]{2\cdot 3}\,\sqrt[3]{3\cdot 5} + \sqrt[3]{3\cdot 5}\,\sqrt[3]{5\cdot 2} + \sqrt[3]{5\cdot 2}\,\sqrt[3]{2\cdot 3}) + 3 z^2 \\ & = y^3 -3y \sqrt[3]{2\cdot 3\cdot 5}\,(\sqrt[3]{2} + \sqrt[3]{3} + \sqrt[3]{5}) + 3 z^2 \\ & = y^3-3xyz + 3 z^2 \tag{6} \end{align}

The rest of the derivation consists of eliminating $y$ between $(5),(6)$ to obtain an algebraic equation in $x,z$, then eliminating $z$ between that equation and $z^3=30$ to obtain the final polynomial in $x$.

From $(5)\,$:

$$y = \frac{x^3+3z-10}{3x} \tag{7}$$

Substituting the expression for $y$ from $(7)$ in $(6)\,$ and using $z^3=30\,$ from $(3)\,$ at the last step:

\require{cancel} \begin{align} 0 & = y^3-3xyz + 3 z^2 - 31 \\ & = \frac{(x^3+3z-10)^3}{27 x^3} - \cancel{3x} \frac{x^3+\bcancel{3z}-10}{\cancel{3x}} z + \bcancel{3 z^2} - 31 \\ \iff 0 & = (x^3+3z-10)^3- 27 x^3(x^3-10)-27\cdot 31 x^3 \\ & = (x^3-10)^3+3(x^3-10)^2z + 27(x^3-10)z^2+27z^3- 27 x^3(x^3-10)-27\cdot 31 x^3 \\ & = 27(x^3-10)\,z^2-9(2x^6-10x^3-100)\,z+(x^3-10)^3-27\,(31 x^3-30) \tag{8} \end{align}

Define the polynomials in $x\,$:

$$\begin{cases} u = 27(x^3-10) \\ v = 9 (2 x^6 - 10 x^3 - 100) = 18(x^3-10)(x^3+5) \\ w = (x^3-10)^3-27\,(31 x^3-30) = x^9 - 30 x^6 - 537 x^3 - 190 \tag{9} \end{cases}$$

Then $(8)$ becomes:

$$u z^2 = v z - w \tag{10}$$

Raising to the $3^{rd}$ power and again using $z^3=30\,$:

\begin{align} 900 u^3 = u^3 z^6 & = (vz-w)^3 \\ & = v^3z^3 - w^3 -3vz w \cdot (vz-w) \\ & = 30 v^3 - w^3 -3vwz \cdot u z^2 \\ & = 30 v^3 -w^3 - 90 uvw \tag{11} \\ \iff \quad w^3 + 90 uvw - 30 v^3 + 900 u^3 & = 0 \tag{12} \end{align}

Given that $w$ is monic and $\deg u = 3, \deg v = 6, \deg w = 9$ as polynomials in $x\,$, $(12)$ defines a monic polynomial of degree $27$ with integer coefficients having $x = \sqrt[3]{2} + \sqrt[3]{3} + \sqrt[3]{5}$ as a root.

For verification, one can substitute the expressions for $u,v,w$ from $(9)$ into $(12)$ and derive the actual polynomial in $x$ after routine, albeit not pretty, algebraic manipulations...

$$x^{27} - 90 x^{24} + 1089 x^{21} - 62130 x^{18} + 105507 x^{15} - 16537410 x^{12} - 30081453 x^9 - 1886601330 x^6 + 73062900 x^3 - 6859000 \tag{13}$$

...then verify that $x = \sqrt[3]{2} + \sqrt[3]{3} + \sqrt[3]{5}$ is indeed a root.

As a side note, the same technique can be used to find a monic polynomial with integer coefficients having $x=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$ as a root for arbitrary $a,b,c \in \mathbb{Z}$. However, in the case here the polynomial $(13)$ happens to be the minimal polynomial for the given value, but that's not necessarily true in general, for example for $x=\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{-3}$ or $x=\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{6}\,$.

Hint: Suppose you want $\sqrt{2}+\sqrt3$ as the root of a polynomial with integer coefficients. Then we have:

\begin{align} x&=\sqrt2+\sqrt3\\ x^2&=5+2\sqrt6\\ x^2-5&=2\sqrt6\\ (x^2-5)^2&=24\\ (x^2-5)^2-24&=0. \end{align} Thus, $\sqrt2+\sqrt3$ is a root of $$f(x)=x^4-10x^2+1.$$

• Beware : it is cubic roots... Dec 8, 2016 at 18:33
• @hkmather802 This is a hint, not the answer so Clayton isn't going to provide the exact solution. Dec 8, 2016 at 18:34
• Guys, don't down vote the hint. It's the way to find the answer without explicitly giving the answer away. Dec 8, 2016 at 18:34
• But it is just an easy case that one is cubic root with three radicals. Dec 8, 2016 at 18:36
• This approach is much less efficient with cubics... Dec 8, 2016 at 18:47

Let $u=\cos(2\pi/3)+i\sin(2\pi/3)$. This is a third primitive root of unity, $u^3=1$. Next, let $a=\sqrt[3]2$, $b=\sqrt[3]3$, and $c=\sqrt[3]5$. The desired polynomial can be taken in the form $$P(x)\equiv P(x,a,b,c)=\prod_{j,k,l=1}^3(x-u^ja-u^kb-u^lc)$$ and has degree $27$ (because there are $3\times3\times3=27$ factors in the product). Indeed,

$\bullet$ It is clearly monic.

$\bullet$ If we replace $a$ by $ua$ in the definition of the polynomial, then it does not change (the factors are just permuted, $a\to ua$, $ua\to u^2a$, $u^2a\to u^3a=a$). On the other hand, the coefficient of $a^s$ in $P(x,a,b,c)$ is multiplied by $u^s$. Thus, if $s$ is not a multiple of $3$, then the coefficient of $a^s$ is zero. In other words, only $a^3=2$, $a^6=4$ etc. occur in the coefficients. The same is true of $b$ and $c$. Thus, the coefficients of $P(x)$ are sums of products of integers by powers of $u$.

$\bullet$ The coefficients of the polynomial $P(x)$ are real (because complex conjugation just interchanges $u$ and $u^2$). Let $R$ be any of the coefficients. By the preceding, $R=m+nu+pu^2$, where $m,n,$ and $p$ are integers; since $R$ is real, we have $p=n$, and $R=m+n(u+u^2)=m-n$ is an integer. Thus, the polynomial has integer coefficients.

$\bullet$ Clearly, $a+b+c$ is a root of $P(x)$.

Thus, $P(x)$ satisfies all the necessary requirements.