Question-100561

Question Number 100561 by DGmichael last updated on 27/Jun/20

Commented by bramlex last updated on 27/Jun/20

x - y = \frac{x^{3} - y^{3}}{x^{2} + x y + y^{2}}

x = \sqrt[3]{3 + \sqrt{9 + \frac{125}{27}}}, y = \sqrt[3]{- 3 + \sqrt{9 + \frac{125}{27}}}

x - y = \frac{6}{\sqrt[3]{{(3 + \sqrt{9 + \frac{125}{27}})}^{2}} + \sqrt[3]{\frac{125}{27}} + \sqrt[3]{{(- 3 + \sqrt{9 + \frac{125}{27}})}^{2}}}

= \frac{6}{\frac{5}{3} + \sqrt[3]{18 + \frac{125}{27} + 6 \sqrt{9 + \frac{125}{27}}} + \sqrt[3]{18 + \frac{125}{27} - 6 \sqrt{9 + \frac{125}{27}}}}

Answered by MJS last updated on 27/Jun/20

A is the solution of A^{3} + p A + q = 0

Cardano

A = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}} =

= \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}} - \sqrt[3]{\frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}

\frac{q}{2} = 3 \Rightarrow q = 6

\frac{q^{2}}{4} = 9 \Rightarrow q = \pm 6

\Rightarrow q = 6

\frac{p^{3}}{27} = \frac{125}{27} \Rightarrow p = 5

A^{3} + 5 A + 6 = 0

\Rightarrow A_{1} = - 1 \land A_{2, 3} = \frac{1}{2} \pm \frac{\sqrt{23}}{2} i

\Rightarrow A = A_{1} = - 1

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