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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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Did you have any luck with this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0solve for K K = p / (3H) substitute into H^3 + K^3 = q H^3 + (p / (3H))^3 = q Can you continue from here?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1@concad I have noticed you have done this with previous problems. Close questions and then make a new question with the same question. Is this because you don't want to do the problem and you want us to? This isn't how OpenStudy is suppose to work. We are suppose to be here to help, not do your homework for you.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1@jayzdd has actually did exactly what the problem has asked up till the solving for H part he solved that one equation for K and plugged in that result into the other equation I gave you another hint on the previous thread...I said you should expect a quadratic in terms of H^3.
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