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Problem

Rationalize $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$

So, I'm training for the Mexican Math Olimpiad. This is one of the algebra problems from my weekly training. Before this problem, there was other very similar, after proving it, there's an useful property:

If $a+b+c=0$ then $a^3+b^3+c^3=3abc$

It can be easily proved using the following factorization

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$

I tried using this one for the rationalization. I got

$$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}*\frac{(a^2+b^2+c^2-ab-bc-ac)}{(a^2+b^2+c^2-ab-bc-ac)}=\frac{\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}-\sqrt[3]{ab}-\sqrt[3]{bc}-\sqrt[3]{ac}}{a+b+c-3\sqrt[3]{abc}}$$

But i didn't know how to proceed. I tried looking into it with wolfram alpha and i got there is no racionalization, so i assume $a+b+c=0 $ (not $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$, because the first expression wouldn't have sense) if we want a solution (In fact, if this happens, i've already rationalized it).

So my truly question is not the answer, but how to prove that $a+b+c$ needs to be 0

I would appreciate any help, ideas or suggestions, thanks.

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    $\begingroup$ There is no reason at all for $a+b+c$ to be $0$. $\endgroup$ – Robert Israel Jun 3 at 23:29
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In M2:

R=QQ[x,y,z,a,b,c]
I=ideal(x^3-a,y^3-b,z^3-c,x+y+z)
gens gb I
toString oo -- note a^3+3*a^2*b+3*a*b^2+b^3+3*a^2*c-21*a*b*c+3*b^2*c+3*a*c^2+3*b*c^2+c^3
-- writing x^3 for a, y^3 for b and z^3 for c:
factor(x^9+3*x^6*y^3+3*x^6*z^3+3*x^3*y^6-21*x^3*y^3*z^3+3*x^3*z^6+y^9+3*y^6*z^3+3*y^3*z^6+z^9) 
tex oo

$\left(x+y+z\right)\left(x^{2}-x\,y-x\,z+y^{2}-y\,z+z^{2}\right)\left(x^{2}-x\,y-x\,z+y^{2}+2\, y\,z+z^{2}\right)\left(x^{2}-x\,y+2\,x\,z+y^{2}-y\,z+z^{2}\right)\left(x^{2}+2\,x\,y-x\,z+y^{2 }-y\,z+z^{2}\right)$

Which means $\frac1{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}=\frac1{x+y+z}=\frac{(x^2-xy-xz+y^2-yz+z^2)(x^2-xy-xz+y^2+2yz+z^2)(x^2-xy+2xz+y^2-yz+z^2)(x^2+ 2xy-xz+y^2-yz+z^2)}{x^9+3x^6y^3+3x^6z^3+3x^3y^6-21x^3y^3z^3+3x^3z^6+y^9+3y^6z^3+3y^3z^6+z^9}=\frac{({\sqrt[3]{a}}^2-{\sqrt[3]{a}}{\sqrt[3]{b}}-{\sqrt[3]{a}}{\sqrt[3]{c}}+{\sqrt[3]{b}}^2-{\sqrt[3]{b}}{\sqrt[3]{c}}+{\sqrt[3]{c}}^2)({\sqrt[3]{a}}^2-{\sqrt[3]{a}}{\sqrt[3]{b}}-{\sqrt[3]{a}}{\sqrt[3]{c}}+{\sqrt[3]{b}}^2+2{\sqrt[3]{b}}{\sqrt[3]{c}}+{\sqrt[3]{c}}^2)({\sqrt[3]{a}}^2-{\sqrt[3]{a}}{\sqrt[3]{b}}+2{\sqrt[3]{a}}{\sqrt[3]{c}}+{\sqrt[3]{b}}^2-{\sqrt[3]{b}}{\sqrt[3]{c}}+{\sqrt[3]{c}}^2)({\sqrt[3]{a}}^2+ 2{\sqrt[3]{a}}{\sqrt[3]{b}}-{\sqrt[3]{a}}{\sqrt[3]{c}}+{\sqrt[3]{b}}^2-{\sqrt[3]{b}}{\sqrt[3]{c}}+{\sqrt[3]{c}}^2)}{a^3+3a^2b+3a^2c+3ab^2-21abc+3ac^2+b^3+3b^2c+3bc^2+c^3}$

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Maple rationalizes it as $$ -{\frac { \left( {x}^{2/3}+2\,\sqrt [3]{x}\sqrt [3]{y}-\sqrt [3]{z} \sqrt [3]{x}+{y}^{2/3}-\sqrt [3]{z}\sqrt [3]{y}+{z}^{2/3} \right) \left( 3\,{x}^{5/3}\sqrt [3]{y}-6\,{x}^{2/3}{y}^{4/3}+3\,{x}^{2/3} \sqrt [3]{y}z-6\,{x}^{4/3}{y}^{2/3}+3\,\sqrt [3]{x}{y}^{5/3}+3\,\sqrt [3]{x}{y}^{2/3}z-{x}^{2}+7\,xy-2\,xz-{y}^{2}-2\,yz-{z}^{2} \right) }{{ x}^{3}+3\,{x}^{2}y+3\,{x}^{2}z+3\,x{y}^{2}-21\,xyz+3\,x{z}^{2}+{y}^{3} +3\,{y}^{2}z+3\,y{z}^{2}+{z}^{3}}} $$

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  • $\begingroup$ That's quite a lot. I'll try to get this answer, thank you $\endgroup$ – SonodaUmi Jun 4 at 3:32
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You can proceed by using $x^3-y^3=(x-y)(x^2+xy+y^2).$

If $a+b+c=0$ and $a^2+b^2+c^2-ab-ac-bc=0$ so the last gives $$(a-b)^2+(a-c)^2+(b-c)^2=0$$ or $a=b=c,$ which gives $a=b=c=0,$ which is impossible.

Thus, your trick works for all reals $a$, $b$ and $c$ such that $$a+b+c\neq0$$ and $$(a-b)^2+(a-c)^2+(b-c)^2\neq0,$$ which says that at least two numbers from $\{a,b,c\}$ must be different.

I changed $\sqrt[3]a$ by $a$ and similar, but the essence does not change.

If $a=b=c$ so we work with $$\frac{1}{3\sqrt[3]a},$$ which is $$\frac{\sqrt[3]{a^2}}{3a}.$$

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  • $\begingroup$ That's a very good idea, i should have tried to prove it before saying it. Thank you $\endgroup$ – SonodaUmi Jun 4 at 3:33
  • $\begingroup$ @greatpulie You are welcome! $\endgroup$ – Michael Rozenberg Jun 4 at 4:17

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