Cube roots fun

Cube both sides and factor to get: $18 - 3\sqrt[3]{(x+9)(x-9)}(\sqrt[3]{x+9} - \sqrt[3]{x-9}) = 27.$

Simplifying and substituting 3 for $\sqrt [3]{x + 9} - \sqrt [3]{x - 9}$ , we get: $- \sqrt [3]{x^2 - 81} = 1.$

From here, we simply cube both sides and add 81 to get $x^2 = \boxed{80}$ .

Let $\space \begin{cases} x+9 = (a+b)^3 = a^3+3a^2b+3ab^2+b^3 &...(1) \\ x-9 = (a-b)^3 = a^3-3a^2b+3ab^2-b^3 &...(2) \end{cases}$

$\Rightarrow \sqrt [3] {x+9} - \sqrt [3] {x-9} = \sqrt [3] {(a+b)^3} - \sqrt [3] {(a-b)^3} = (a+b)-(a-b) \\ \quad = 2b = 3 \quad \Rightarrow b = \dfrac {3}{2}$

Eq. 1 - Eq. 2: $\space 6a^2b+2b^3 = 18 \quad \Rightarrow 9a^2 + \dfrac {27}{4} = 18 \quad \Rightarrow a = \pm \dfrac {\sqrt{5}}{2}$

From Eq. 1:

$\begin{aligned} x+9 & = (a+b)^3=\left( \dfrac {\pm \sqrt{5}+3}{2} \right)^3 = \dfrac {\pm 5\sqrt{5}+45\pm27\sqrt{5}+27}{8} \\ & = \dfrac {\pm 32\sqrt{5}+72}{8} = \pm 4\sqrt{5}+9 \end{aligned}$

$\Rightarrow x = \pm 4\sqrt{5}\quad \Rightarrow x^2 = \boxed{80}$

My answer could be a little longer than the others, but I consider it's still functional:

Let's call $\sqrt [ 3 ]{ x+9 } =a$ and $\sqrt [ 3 ]{ x-9 } =b$ .

So, the equation can be rewritten: $a-b=3$ .

Considering the fact that we're dealing with cubic roots, I used the factorization $\left( a-b \right) \left( { a }^{ 2 }+ab+{ b }^{ 2 } \right) ={ a }^{ 3 }-{ b }^{ 3 }$ .

For the sake of simplicity and clarity, I'll define $F={ a }^{ 2 }+ab+{ b }^{ 2 }$ .

Multiplying both sides of equation by F, we obtain: ${ a }^{ 3 }-{ b }^{ 3 }=3F$ .

The LHS can be simplified as follows: ${ a }^{ 3 }-{ b }^{ 3 }=\left( x+9 \right) -\left( x-9 \right) =18$ .

Then, the equation reduces to: $F=6$ .

We can stablish the non-linear two-equation system: $\begin{cases} a-b=3 \\ { a }^{ 2 }+ab+{ b }^{ 2 }=6 \end{cases}$

There are many ways to solve it. The way I implemented was to rewrite conveniently the second equation, so it's possible substitute the first one into it:

${ a }^{ 2 }-2ab+{ b }^{ 2 }+3ab=6$

${ \left( a-b \right) }^{ 2 }+3ab=6$

$3ab=-3\quad \Rightarrow \quad ab=-1$

Writing back the expressions of a and b, and multiplying the roots, we obtain:

$\sqrt [ 3 ]{ { x }^{ 2 }-81 } =-1\quad \Rightarrow \quad { x }^{ 2 }-81=-1\quad \Rightarrow \quad \boxed { { x }^{ 2 }=80 }$

Hope you enjoy this answer!

NOTE: This is my first contribution to this great community. Please, don't be so rude! :)

Our solution will require very little computation.

Recall the cubic formula: The solutions of $t^3+3qt-2r=0$ are $t=\sqrt[3]{\sqrt{q^3+r^2}+r}-\sqrt[3]{\sqrt{q^3+r^2}-r}$ , which is the "format" of the LHS of the given equation, with $r=9$ and $x=\sqrt[3]{q^3+r^2}$ .

Our idea is to find the cubic $f(t)=t^3+3qt-18$ with $f(3)=0$ . We see that $q=-1$ , and $f(t)=t^3-3t-18$ . Now the cubic equation gives $\sqrt[3]{\sqrt{80}+9}-\sqrt[3]{\sqrt{80}-9}=3$ , so that $x=\sqrt{80}$ and $x^2=80$ .

WE SIMPLY whole cube both side getting 27 on the RHS..then (x+9)-(x-9)-3.cuberoot(x square - 81).(3)....as "a-b=3" see carefully..so 18 -{9.cube root (x square -81)} = 27...on simplifying -1=cube root(x square -81)......-1=x^2 - 81..........x^2=80...is ur ans

we can rewrite this equation as (x+ 9)^ 1/3 = 3 + (x-9)^ 1/3 now we cube both sides , getting X+9 = 27+x -9 +3.3 (( x -9) ^ 1/3) ( 3 + ( x-9 ) ^ 1/3)) we can substitute the left side of the 1st equation and cancel Which are needed to get 9+9 ( x ^2-81)^ 1/3 =0 cancelling 9 we get (x^2-81)^ 1/3 = -1. now we can cube and solve for x^2 to get x^2 = 80