Creative Question I

$\large\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}=\sqrt[3]{x\ \sqrt[3]{x\ \sqrt[3]{x\cdots}}}=k$

Let's solve the second part: $\begin{aligned} \sqrt[3]{x\ \sqrt[3]{x\ \sqrt[3]{x\cdots}}} &= k \\ \sqrt[3]{xk}&= k \\ xk &= k^{3} \\ k &= \sqrt{x} \ \ \ (\because k > 0) \\ \end{aligned}$

Now we back to the first part: $\begin{aligned} \sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}} &= k \\ \sqrt[3]{x+k} &=k \\ \sqrt[3]{x+\sqrt{x}} &= \sqrt{x} \\ x+\sqrt{x} &= x\sqrt{x} \\ \frac{x}{x-1} &=\sqrt{x} \\ \frac{x^{2}}{x^{2}-2x+1} &= x \\ x^{2}-3x+1 &= 0 \ \ \ (\because x \neq 0) \\ \end{aligned}$

By using the formula, we can obtain the solution of $x$ : $\large x=\frac{3+\sqrt{5}}{2} \ \ \ \bigg(\because\frac{x}{x-1}>0 \bigg)$

Hence, we found that $a=9$ , $b=5$ , and the value of $a+b$ is equal to $\large\boxed{14}$ .

Employing the infinite geometric series, note that:

$\sqrt[3]{x\ \sqrt[3]{x\ \sqrt[3]{x\cdots}}}=x^{1/3+(1/3)^2+...}=x^{\frac{1/3}{1-1/3}}=x^{1/2}=\sqrt{x}$

Then we have:

$\ \sqrt[3]{x+\sqrt{x}}=\sqrt{x}$

$(x+\sqrt{x})^{2}=x^3$

$(\sqrt{x}(\sqrt{x}+1))^{2}=x^{1+2}$

$\cancel{x}(\sqrt{x}+1)^2=\cancel{x}x^2$

$(\sqrt{x}+1)^2=x^2$

$\sqrt{x}+1=x$ where we reject $\sqrt{x}+1=-x$ considering that this would make either $x$ or $-\sqrt{x}$ negative.

$\sqrt{x}=x-1$

$x=x^2-2x+1$

$x^2-3x+1=0$

$x=\dfrac{3+\sqrt{3^2-4(1)}}{2}=\dfrac{\sqrt{9}+\sqrt{9-4}}{2}=\dfrac{\sqrt{9}+\sqrt{5}}{2}$

As required, $a+b=9+5=\boxed{14}$ .

Set LHS and RHS = y, then

(1) (x+y)=y^3

(2) y^3=xy

Elliminating y, we get x=1+√x, and it gives the solution as x=(1/2)(√9 +√5), thus, a=9 & b=5 and Answer=(a+b)=14.

Let $u = \sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}} = \sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\cdots}}}$ .

Then

$\begin{aligned} u & = \sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}} & \small \color{#3D99F6} \text{Cubing both sides} \\ \implies u^3 & = x + u & ...(1) \end{aligned}$

And

$\begin{aligned} u & = \sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\cdots}}} & \small \color{#3D99F6} \text{Cubing both sides} \\ u^3 & = xu & \small \color{#3D99F6} \text{For }u \ne 0 \\ \implies x & = u^2 & ...(2) \end{aligned}$

Substitute $x=u^2$ in equation (1):

$\begin{aligned} u^3 & = u^2 + u & \small \color{#3D99F6} \text{For }u \ne 0 \\ u^2-u-1 & = 0 & \small \color{#3D99F6} \text{Solving the quadratic for }u \\ \implies u & = \frac {1\pm \sqrt 5}2 \end{aligned}$

From equation (2): $x = u^2 = \left(\dfrac {1\pm \sqrt 5}2\right)^2 = \begin{cases} \dfrac {6+2\sqrt 5}4 = \dfrac {3+\sqrt 5}2 = \dfrac {\sqrt 9+\sqrt 5}2 \\ \dfrac {6+2\sqrt 5}4 = \dfrac {3-\sqrt 5}2 \quad \color{#D61F06} \text{Rejected (see note)}\end{cases}$

Therefore, $a+b = 9 + 5 = \boxed{14}$ .

Note: Putting $x = \frac {3-\sqrt 5}2$ in $\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}} \approx 1.153721376$ , but putting it in $\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\cdots}}} \approx 0.618033989$ .

Let the given expression = A and cube

we have $A^3 = x+A = xA$

This gives $x+A = xA$

$x = xA - A\\ x = A(x-1)\\ \frac{x}{(x-1)} = A$

Since $A^3 = xA$ we have

$\frac{x^3}{(x-1)^3} = x\frac{x}{x-1}\\ \frac{x^3}{(x-1)^3} = \frac{x^2}{x-1}\\$

Since $a$ and $b$ are integers, $x \neq{1}$ and $x - 1 \neq{0}$

Cancelling

$\frac{x}{(x-1)^2} = 1\\ x = (x-1)^2\\ x = x^2 - 2x +1\\ x^2 - 3x +1 = 0$

Using the general solution and solving for x:

$x = \frac{3 + \sqrt{5}}{2}\\ x = \frac{\sqrt{9} + \sqrt{5}}{2}$

since both a and b are to be positive integers

Thus $a = 9$ and $b = 5$ so $a + b = 14$

Substitution is everything: $\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}=y$ and $\sqrt[3]{x\ \sqrt[3]{x\ \sqrt[3]{x\ldots}}}=z$ .

We can then write $y=\sqrt[3]{x+y}$ and $z=\sqrt[3]{xz}$ .

Cube both equations and solve for $x$ : $x=y^{3}-y$ and $x=z^{2}$ . So, $y^{3}-y=z^{2}$ .

Remember $y=z$ . Thus, $y^{3}-y^{2}-y=0$ . Solving for $y$ , we get $y=\frac{1+\sqrt{5}}{2}$ .

Since $y=z$ , just square $y$ to find $x$ : $x=\frac{\sqrt{9}+\sqrt{5}}{2}$ . So the answer is $9+5=\boxed{14}$

$\sqrt [3]{x + \sqrt [3]{x+...}} = \sqrt [3]{x \sqrt [3]{x...}}$

x + $\sqrt [3]{x + \sqrt [3]{x+...}} = x \sqrt [3]{x \sqrt [3]{x...}}$

x + $\sqrt [3]{x \sqrt [3]{x...}}$ = x $\sqrt [3]{x \sqrt [3]{x...}}$

x= $\sqrt [3]{x \sqrt [3]{x...}}$ (x-1)

$x^3 =x \sqrt [3]{x \sqrt [3]{x...}} (x-1)^3$

$x^3 = x (x-1)^2 \sqrt [3]{x \sqrt [3]{x...}} (x-1)$

$x^3 = x (x-1)^2 x$

$x^2 [ (x-1)^2 -x]$ =0

$x^2 =0$ , or $x^2 -3x +1=0$

Because x > 0

x= $\frac {3 + \sqrt {9-4}} {2}$

x = $\frac {\sqrt {9} + \sqrt {5}} {2} =\frac { \sqrt {a} + \sqrt {b}} {2}$

So a + b = 14