When I solve the Project Euler #251. The formula $$ \sqrt[3]{a+b\sqrt{c}} + \sqrt[3]{a-b\sqrt{c}} = 1 $$ could be transformed into $$ 8a^3+15a^2 + 6a -27b^2c =1 $$ But I didn't sort it out.
How to do the transformation?
Cardano Triplet transformation.
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linear-algebra
algebra-precalculus
1 Answers
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We have, $$\sqrt[3]{a+b\sqrt{c}} =1-\sqrt[3]{a-b\sqrt{c}} \tag{1}$$ $$\Rightarrow a +b\sqrt{c} = 1-3\sqrt[3]{a-b\sqrt{c}}[1-\sqrt[3]{a-b\sqrt{c}}] -a +b\sqrt{c} \,\,\,\,(\text{by cubing (1)})$$ $$\Rightarrow 2a-1 =- 3\sqrt[3]{a-b\sqrt{c}}[1-\sqrt[3]{a-b\sqrt{c}}] =- 3\sqrt[3]{a-b\sqrt{c}}\sqrt[3]{a+b\sqrt{c}}\,\,\,(\text{using (1) to simplify})\tag{2}$$ $$\Rightarrow 8a^3-12a^2+6a-1 = -27(a^2-b^2c)\,\,\,\,(\text{by cubing (2)})$$ $$\Rightarrow \boxed{8a^3+15a^2+6a-27b^2c = 1}$$