Express $\sqrt[3]{(7+5\sqrt{2})}$ in the form $x+y\sqrt{2}$ with $x$ and $y$ rational numbers.

I.e. Show that it is $1+\sqrt{2}$.


closed as off-topic by Watson, Roman83, Alex Mathers, user133281, Davide Giraudo Sep 25 '16 at 21:09

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    $\begingroup$ Related: math.stackexchange.com/questions/1178996, math.stackexchange.com/questions/790738. $\endgroup$ – Watson Sep 25 '16 at 13:10
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    $\begingroup$ You could just cube $(1+\sqrt2)$ and be done with it :-) Your number passes the lithmus test of norm: $N(7+5\sqrt2)=7^2-5\cdot2^2=-1$, which is a cube, so the number may be a cube itself. If a nice cube root exists, it has to be pretty small, so trial and error should be ok. Bill Dubuque has described his denesting algorithm here in many threads, but IIRC that is for square roots of quadratic integers. Not sure whether there is a variant for cube roots? $\endgroup$ – Jyrki Lahtonen Sep 25 '16 at 13:15
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    $\begingroup$ i would calculate $$(1+\sqrt{2})^3$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 25 '16 at 13:18
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    $\begingroup$ @JyrkiLahtonen I wrote math.stackexchange.com/questions/396915/… a while ago addressing this. $\endgroup$ – mercio Sep 25 '16 at 14:05
  • $\begingroup$ A good one, @mercio! Thanks for adding the link. $\endgroup$ – Jyrki Lahtonen Sep 25 '16 at 14:09

You can assume that the nested radical can be expressed in $a+b\sqrt{2}$ form. More specifically, we have $$\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}\tag{1}$$ With your question, we have $$\sqrt[3]{7+5\sqrt{2}}=a+b\sqrt{2}\tag{2}$$

Cubing both sides, we get $$7+5\sqrt{2}=(a^3+6ab^2)+(3a^2b+2b^3)\sqrt{2}\tag{3}$$ And equating corresponding coefficients, we get the following system of equations: $$\begin{cases}a^3+6ab^2=7\\3a^2b+2b^3=5\tag{4}\end{cases}$$ Cross multiplying, we get a multi-variate polynomial. Namely, $$5a^3-21a^2b+30ab^2-14b^3=0\tag{5}$$ Dividing both sides by $b^3$, we get: $$5\frac {a^3}{b^3}-21\frac {a^2}{b^2}+30\frac {a}{b}-14=0\tag{6}$$ Which is also equal to $5\left(\frac ab\right)^3-21\left(\frac {a}{b}\right)^2+30\left(\frac {a}{b}\right)-14=0$. Substituting $a/b$ with $x$, we get the cubic polynomial$$5x^3-21x^2+30x-14=0\tag{7}$$ with $x=1$ as an integer root. Since $a/b=x$, we have $$\frac ab=1\implies a=b\tag{8}$$ So from $(3)$, we have $a^3+6a(a)^2=7\implies a^3+6a^3=7\implies 7a^3=7\implies a=b=1$


For practice, you can try to denest $\sqrt[3]{2+\sqrt{5}}$

  • $\begingroup$ The fourth and third line from the bottom is confusing. I don't know what you did to get a=b. Could you explain further when solving line (5). $\endgroup$ – James257 Sep 25 '16 at 15:58
  • $\begingroup$ @James257 In $(4)$, we have $5(a^3+6ab^2)=7(3a^2b+2b^3)$. Expanding and moving all terms to the left, we get $(5)$. Divide both sides by $b^3$ to get $(6)$ and set $x=a/b$ so we get $$5x^3−21x^2+30x−14=0$$ with root $x=1$. But we want $a$ and $b$ so we replace $x$ with its substitution $\frac ab$ to obtain $\frac ab=1\implies a=b$ $\endgroup$ – Frank Sep 25 '16 at 17:48

$$(1+\sqrt{2})^3=(1+2\sqrt{2}+2)(1+\sqrt{2})=(3+2\sqrt{2})(1+\sqrt{2})=3+3\sqrt{2}+2\sqrt{2}+4$$ $$\therefore (1+\sqrt{2})^3=7+5\sqrt{2}$$ $$\Rightarrow 1+\sqrt{2}=\sqrt[3]{7+5\sqrt{2}}$$


If it simplifies, then $7+5\sqrt 2$ is a cube $(a+b \sqrt 2)^3$, in the ring of integers of $\Bbb Q(\sqrt 2)$, which is $\Bbb Z[\sqrt 2]$, so $a$ and $b$ must be integers (sometimes you can only deduce that $2a,a+b,2b$ are integers but it's still very good)

Moreover, you have $2a = (7+5\sqrt 2)^\frac 13 + (7-5\sqrt 2)^\frac 13$

Since $5\sqrt 2$ is between $7$ and $8$, The first term is between $2$ and $3$ the second term is between $-1$ and $0$, so the sum has to be $2$ if it's going to be an even integer. So we can bet on $a=1$.

Then writing $7+5\sqrt 3 = (1+b\sqrt 2)^3$ you get $7 = 1+6b^2$, thus $b^2=1$, and you also get $5 = 3b+2b^3 =b(3+2b^2) = b(3+2) = 5b$ so $b=1$.

Since it is compatible with $b^2=1$, it shows that $(1+\sqrt 2)^3 = 7+5\sqrt 2$


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