- #1

Math100

- 779

- 220

- Homework Statement
- Prove without using Cardano's formula that ## \sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}}=3 ##.

- Relevant Equations
- None.

Proof:

Let ## x=\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}} ##.

Then ## x^3=(\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}})^3 ##.

Note that ## (a+b)^3=a^3+3ab(a+b)+b^3 ## where ## a=\sqrt[3]{18+\sqrt{325}} ## and ## b=\sqrt[3]{18-\sqrt{325}} ##.

This gives ## x^3=(\sqrt[3]{18+\sqrt{325}})^3+3\sqrt[3]{(18+\sqrt{325})(18-\sqrt{325})}\cdot (\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}})+(\sqrt[3]{18-\sqrt{325}})^3\implies x^3=18+\sqrt{325}+3\sqrt[3]{(18+\sqrt{325})(18-\sqrt{325})}\cdot x+18-\sqrt{325}\implies ##.

Now we have ## x^3=36+3\sqrt[3]{(18+\sqrt{325})(18-\sqrt{325})}\cdot x\implies x^3=36+3\sqrt[3]{-1}\cdot x\implies x^3=36-3x ##,

so ## x^3+3x=36 ##.

From here, it's clear and obvious that ## x=3 ## is a solution since ## 3^3+3(3)=36 ## by direct substitution. But how to solve this cubic equation ## x^3+3x=36 ## so that I can get ## x=3 ## to finish this proof?

Let ## x=\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}} ##.

Then ## x^3=(\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}})^3 ##.

Note that ## (a+b)^3=a^3+3ab(a+b)+b^3 ## where ## a=\sqrt[3]{18+\sqrt{325}} ## and ## b=\sqrt[3]{18-\sqrt{325}} ##.

This gives ## x^3=(\sqrt[3]{18+\sqrt{325}})^3+3\sqrt[3]{(18+\sqrt{325})(18-\sqrt{325})}\cdot (\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}})+(\sqrt[3]{18-\sqrt{325}})^3\implies x^3=18+\sqrt{325}+3\sqrt[3]{(18+\sqrt{325})(18-\sqrt{325})}\cdot x+18-\sqrt{325}\implies ##.

Now we have ## x^3=36+3\sqrt[3]{(18+\sqrt{325})(18-\sqrt{325})}\cdot x\implies x^3=36+3\sqrt[3]{-1}\cdot x\implies x^3=36-3x ##,

so ## x^3+3x=36 ##.

From here, it's clear and obvious that ## x=3 ## is a solution since ## 3^3+3(3)=36 ## by direct substitution. But how to solve this cubic equation ## x^3+3x=36 ## so that I can get ## x=3 ## to finish this proof?