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If:$$\sqrt[3]{(x + 9)} - \sqrt[3]{(x-9)} = 3$$ Find $x^2$

I can't seem to solve this question. Any hints or solutions is welcomed.

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  • $\begingroup$ Writing radical with the same power is unusual, do you type truly.? $\endgroup$ – Nosrati Feb 3 '17 at 13:26
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    $\begingroup$ Yes, that's the main part of the confusion. $\endgroup$ – Nia F. Feb 3 '17 at 13:27
  • $\begingroup$ My mistake. power is $\dfrac{1}{3}$. $\endgroup$ – Nosrati Feb 3 '17 at 13:28
  • $\begingroup$ Mathematica says it has not answer. $\endgroup$ – Nosrati Feb 3 '17 at 13:31
  • $\begingroup$ @MyGlasses I think that the OP mean $(\dots)^\frac{1}{3}$ and not $\left(\sqrt{\dots}\right)^\frac{1}{3}$ $\endgroup$ – Jan Feb 3 '17 at 13:32
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We have this:

$$(x + 9)^{\frac{1}{3}} - (x-9)^{\frac{1}{3}} = 3$$

Then: $$(x + 9)^{\frac{1}{3}} -3 = (x-9)^{\frac{1}{3}}$$ $$((x + 9)^{\frac{1}{3}} -3)^3 = ((x-9)^{\frac{1}{3}})^3$$ $$x + 9 -9(x+9)^{\frac{2}{3}} + 27(x+9)^{\frac{1}{3}} -27 = x -9$$ $$-9(x + 9)^{\frac{2}{3}} +27(x +9)^{\frac{1}{3}} =-9-9 +27 $$ $$-9(x + 9)^{\frac{2}{3}} +27(x +9)^{\frac{1}{3}} =9 $$ $$3(x+9)^{\frac{1}{3}} - (x+9)^{\frac{2}{3}} = 1$$ $$(x+9)^{\frac{2}{3}} - 3(x+9)^{\frac{1}{3}} + 1 = 0$$ Let $y = (x+9)^{\frac{1}{3}}$

Then we have: $$y^2 -3y +1 - 0$$ $$y = \frac{3 \pm \sqrt{5}}{2} = (x +9)^{\frac{1}{3}}$$ $$(\frac{72+32\sqrt{5}}{8})^{\frac{1}{3}} = (x + 9)^{\frac{1}{3}}$$ $$\frac{72+32\sqrt{5}}{8} = x + 9$$ $$72 + 32\sqrt{5} = 8x + 72$$ $$8x = 32\sqrt{5}$$ Hence:

$$x = 4\sqrt{5} = \sqrt{80}$$


So $x^2$ would be $80$

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Put $a=(x+9)^{1/3},b=(x-9)^{1/3}$ $$a-b=3\\(a-b)^3=3^3\\a^3-3ab(a-b)+b^3=27$$ Now you already know that $a-b=3$ so $$a^3-9ab-b^3=27$$ Now plugging back $a,b$ $$((x+9)^{1/3})^3-9(x+9)^{1/3}(x-9)^{1/3}-((x-9)^{1/3})^3=27\\x+9-9(x^2-81)^{1/3}-(x-9)=27\\18-9(x^2-81)^{1/3}=27\\(x^2-81)^{1/3}=-1\\x^2-81=-1\\x^2=80$$

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If $(9+x)^\frac 13 + (9-x)^\frac 13 = 3$,
then there is a $y$ such that $(9+x)^\frac 13 = (\frac 32+y)$ and $(9-x)^\frac 13 = (\frac 32-y)$.

Taking cubes of both equations you get $9 \pm x = (\frac {27} 8+\frac 92y^2) \pm (\frac {27} 4 +y^2)y$,
and so $9 = \frac {27} 8+\frac 92y^2$ and $x = (\frac {27} 4 +y^2)y$.

This gives you $y^2 = \frac 54$, then $x = \frac {32} 4 y = 8y$, and so $x^2 = 64y^2 = 16 \times 5 = 80$.

So $(9 + \sqrt {80})^\frac 13 + (9 - \sqrt {80})^\frac 13 = (\frac {3 + \sqrt 5}2) + (\frac {3 - \sqrt 5}2) = 3$ .

As a bonus, you also know the values of the individual cube roots.

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For $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$ we have an interesting problem.

It's $$\sqrt[3]{x+9}+\sqrt[3]{9-x}-3=0$$ and since $\sqrt[3]{x+9}=\sqrt[3]{9-x}=-3$ is impossible, it's $$9+x+9-x-27-9\sqrt[3]{(9+x)(9-x)}=0,$$ which gives $x^2=80$.

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

where $a^2+b^2+c^2-ab-ac-bc=0$ for $a=b=c$ only, which is impossible in our problem.

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  • $\begingroup$ I fixed a typo. $\endgroup$ – Michael Rozenberg Feb 3 '17 at 14:32

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