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anonymous
 one year ago
Historical Feature
3HK=p and H^3+K^3=q
solve for K in 3HK=p and substitute into H^3+K^3=q. Then show that
H=3 sqrt q/2+sqrtq^2/4+p^3/27
anonymous
 one year ago
Historical Feature 3HK=p and H^3+K^3=q solve for K in 3HK=p and substitute into H^3+K^3=q. Then show that H=3 sqrt q/2+sqrtq^2/4+p^3/27

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1Well it says to solve 3HK=p for K then substitute that into H^3+K^3=q have you done this part and if so can you show how far you have gotten with it?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok well if you do that you should end up with a quadratic in terms of H^3
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