Question

How to show

\(\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}=2\)

Answer 1

hmm take cube both side ;;P

Answer 2

so ugly, is this the only way?

Answer 3

o^_^o dunno

Answer 4

btw gtg see y later:)

Answer 5

pz good day

Answer 6

\(\large \begin{align} \color{black}{
x=\sqrt[3]{10+\sqrt{108}} -\sqrt[3]{-10+\sqrt{108}} \hspace{.33em}\\~\\
x=\sqrt[3]{10+6\sqrt{3}} -\sqrt[3]{-10+6\sqrt{3}} \hspace{.33em}\\~\\
x=\sqrt[3]{3\sqrt{3}+9+3\sqrt{3}+1} -\sqrt[3]{3\sqrt{3}-9+3\sqrt{3}-1} \hspace{.33em}\\~\\
x=\sqrt[3]{(\sqrt{3})^3+3\sqrt{3}^2+3\sqrt{3}+1} -\sqrt[3]{(\sqrt{3})^3-3\sqrt{3}^2+3\sqrt{3}-1} \hspace{.33em}\\~\\
x=\sqrt[3]{(\sqrt{3}+1)^3} -\sqrt[3]{(\sqrt{3}-1)^3} \hspace{.33em}\\~\\
x=(\sqrt{3}+1)-(\sqrt{3}-1) \hspace{.33em}\\~\\
x=2
}\end{align}\)

Answer 7

hey @zocker

Answer 8

\[\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}} \\ \sqrt[3]{10+6 \sqrt{3}}+\sqrt[3]{10-6 \sqrt{3}} \\ \\ (a+b \sqrt{3})^3=a^3+ 3 a^2b \sqrt{3}+3 ab^2(3)+b^3 3 \sqrt{3} \\ \text{ gather the irrationals and rationals together } \\ =a^3+9ab^2+3a^2b \sqrt{3}+3 b^3 \sqrt{3} \\ =(a^3+9ab^2)+(3a^2b \sqrt{3}+3 b^3 \sqrt{3}) \\ =(a^3+9ab^2)+(3a^2b +3b^3)\sqrt{3} \\ \text{ so what I'm going to attempt to do is find } a \text{ and } b \text{ such that } \\ a^3+9ab^2=10 \text{ and } 3a^2b+3b^3=6 \]

Answer 9

\[a^2b+b^3=2 \text{ dividing that second equation by } 3\]

Answer 10

2(5)=10...
\[(a^2b+b^3)(5)=a^3+9ab^2 \\ 5a^2b+5b^3=a^3+9ab^2 \\ a^3-5b^3+9ab^2-5a^2b=0 \\ \text{ divide both sides by } b^3 \\ (\frac{a}{b})^3-5+9(\frac{a}{b})-5(\frac{a}{b})^2=0 \\ \text{ Let } x=\frac{a}{b} \\ x^3-5x^2+9x-5=0 \\\]
\[\text{ so } a=b \text{ since } x=1 \\ a^3+9ab^2=10 \\ a^3+9a(a)^2=10 \\ a^3+9a^3=10 \\ 10a^3=10 \\ a=1 \\ b=1 \]

Answer 11

\[\sqrt[3]{10+6 \sqrt{3}}=1+\sqrt{3} \\ \sqrt[3]{10-6 \sqrt{3}}=1-\sqrt{3} \text{ should be able \to show this equality too }\]

Answer 12

\[(1+\sqrt{3})+(1-\sqrt{3}) \\ 2+0\]

Answer 13

wow, this is very impressive on both parts. I gave the medal to the first to answer. I will use both solutions to show my students and give you both props and maybe it will bring more people here ;)

Answer 14

openstudy is so broken
I didn't even see @mathmath333 's response when I posted