Algebra Question

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! :)