MHB Can Cubic Roots and Square Roots Combine to Equal One?

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The discussion revolves around proving that the sum of the cube roots of two expressions, cbrt{2 + sqrt{5}} and cbrt{2 - sqrt{5}}, equals one. Participants suggest raising both sides to the third power and using the identity for the expansion of a binomial cube. The calculations show that both cube roots can be expressed in terms of (1 ± sqrt{5})/2, leading to the conclusion that their sum is indeed 1. The conversation emphasizes solving the problem manually rather than relying on calculators, highlighting the satisfaction of working through the math by hand. The proof is confirmed as correct, reinforcing the validity of the approach.
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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?
 
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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 $$\sqrt[3]{2+ \sqrt{5}}+\sqrt[3]{2- \sqrt{5}}=\frac{1+ \sqrt{5}}{2}+\frac{1- \sqrt{5}}{2}=1$$
 
RTCNTC said:
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
 
mathmari said:
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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kaliprasad said:
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.
 
RTCNTC said:
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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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.
 
Thank you everyone.
 
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