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# Algebra Question

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I found a really interesting problem with a nice solution.

Find the number of real solutions to this equation.

$$\sqrt[3]{x+14}-\sqrt[3]{x-14}=\sqrt[3]{4}$$

I may have remembered the problem wrong  ;-;

Jun 19, 2024

#1
+1075
+1

We can't go ahead and directly solve for x, so let's take a different approach.

Let's first set $$u=\sqrt[3]{x+14}$$. We can solve for u, then plug it in to solve for x.

For this value of u, we get $$x=u^3-14$$

Subsituting out all x, we get that

$$u-\sqrt[3]{\left(u^3-14\right)-14}=\sqrt[3]{4}$$

Removing the roots, we have

$$-u^3+28=4-3\cdot \:4^{\frac{2}{3}}u+3\sqrt[3]{4}u^2-u^3$$

Combining all like terms, we ge tthat

$$3\sqrt[3]{4}u^2-3\cdot \:4^{\frac{2}{3}}u-24=0$$

Using the quadratic equation, we have

$$u=\frac{4^{\frac{2}{3}}+2\left(8\sqrt[3]{4}+4^{\frac{1}{3}}\right)^{\frac{1}{2}}}{2\sqrt[3]{4}},\:u=\frac{4^{\frac{2}{3}}-2\left(8\sqrt[3]{4}+4^{\frac{1}{3}}\right)^{\frac{1}{2}}}{2\sqrt[3]{4}}$$

Now subbing this value of u and finding x, we have

$$\sqrt[3]{x+14}=\frac{4^{\frac{2}{3}}+6\sqrt[3]{2}}{2\sqrt[3]{4}} \\ x+14=\frac{55}{2}+\frac{9}{2}\\ x=18$$

and

$$\sqrt[3]{x+14}=\frac{4^{\frac{2}{3}}-6\sqrt[3]{2}}{2\sqrt[3]{4}}\\ x+14=\frac{1}{2}-\frac{9}{2}\\ x=-18$$

So our answers are 18 and -18.

I completed this in a really complicated manner. I'm interested to see if there was a more efficient way to complete this.

Thanks! :)

Jun 19, 2024