Can Cubic Roots and Square Roots Combine to Equal One?

[mathdad]

Let cbrt = cube root

Let sqrt = square root

Show that
cbrt{2 + sqrt{5}} + cbrt{2 - sqrt{5}} = 1 without using a calculator.

Can someone get me started?

Do I raise both sides to the third power as step 1?

 

[mathmari]

It holds that $$(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3$$

We have the following:
\begin{align*}&\left(1\pm \sqrt{5}\right )^3=1\pm 3\sqrt{5}+3\cdot 5\pm \sqrt{5}^3 =1\pm 3\sqrt{5}+15\pm 5\sqrt{5} =16\pm 8\sqrt{5}=8\left (2\pm \sqrt{5}\right )\\ & \Rightarrow 2\pm \sqrt{5}=\frac{\left(1\pm \sqrt{5}\right )^3}{8} \\ & \Rightarrow \sqrt[3]{2\pm \sqrt{5}}=\sqrt[3]{\frac{\left(1\pm \sqrt{5}\right )^3}{8}} \\ & \Rightarrow \sqrt[3]{2\pm \sqrt{5}}=\frac{1\pm \sqrt{5}}{2}\end{align*}

Therefore we get $$\sqrt[3]{2+ \sqrt{5}}+\sqrt[3]{2- \sqrt{5}}=\frac{1+ \sqrt{5}}{2}+\frac{1- \sqrt{5}}{2}=1$$

 

[kaliprasad]

Let cbrt = cube root

Let sqrt = square root

Show that
cbrt{2 + sqrt{5}} + cbrt{2 - sqrt{5}} = 1 without using a calculator.

Can someone get me started?

Do I raise both sides to the third power as step 1?

you need to prove $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = 1$

you can let $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = x$ and cube both sides and see hat you get after solving it

 

[mathdad]

It holds that $$(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3$$

We have the following:
\begin{align*}&\left(1\pm \sqrt{5}\right )^3=1\pm 3\sqrt{5}+3\cdot 5\pm \sqrt{5}^3 =1\pm 3\sqrt{5}+15\pm 5\sqrt{5} =16\pm 8\sqrt{5}=8\left (2\pm \sqrt{5}\right )\\ & \Rightarrow 2\pm \sqrt{5}=\frac{\left(1\pm \sqrt{5}\right )^3}{8} \\ & \Rightarrow \sqrt[3]{2\pm \sqrt{5}}=\sqrt[3]{\frac{\left(1\pm \sqrt{5}\right )^3}{8}} \\ & \Rightarrow \sqrt[3]{2\pm \sqrt{5}}=\frac{1\pm \sqrt{5}}{2}\end{align*}

Therefore we get $$\sqrt[3]{2+ \sqrt{5}}+\sqrt[3]{2- \sqrt{5}}=\frac{1+ \sqrt{5}}{2}+\frac{1- \sqrt{5}}{2}=1$$

Nicely done! This is not your typical radical equation problem. I could have easily used the wolfram website but this is like cheating. I like to work it out by hand and then check my answer using wolfram or mathway.com.

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you need to prove $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = 1$

you can let $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = x$ and cube both sides and see hat you get after solving it

I understand what you mean but where did x come from? The original equation is equated to 1 not x.

 

[kaliprasad]

Nicely done! This is not your typical radical equation problem. I could have easily used the wolfram website but this is like cheating. I like to work it out by hand and then check my answer using wolfram or mathway.com.

- - - Updated - - -
I understand what you mean but where did x come from? The original equation is equated to 1 not x.

you are supposed to prove that it is 1. you do not know it. so presume that it is x. then cube and remove redicals and solve for x.
it should come to be 1.

 

[mathdad]

Thank you everyone.