Question and Answers Forum

All Questions      Topic List

Logic Questions

Previous in All Question      Next in All Question      

Previous in Logic      Next in Logic      

Question Number 171829 by Mikenice last updated on 21/Jun/22

please i need cubic formula

$${please}\:{i}\:{need}\:{cubic}\:{formula} \\ $$$$ \\ $$

Answered by MathematicalUser2357 last updated on 05/Jan/24

x_1 =−(b/(3a))−(1/(3a))(((2b^3 −9abc+27a^2 d+(√((2b^3 −9abc+27a^2 d)^2 −4(b^2 −3ac)^3 )))/2))^(1/3) −(1/(3a))(((2b^3 −9abc+27a^2 d−(√((2b^3 −9abc+27a^2 d)^2 −4(b^2 −3ac)^3 )))/2))^(1/3) x_2 =−(b/(3a))+((1+i(√3))/(6a))(((2b^3 −9abc+27a^2 d+(√((2b^3 −9abc+27a^2 d)^2 −4(b^2 −3ac)^3 )))/2))^(1/3) −((1−i(√3))/(6a))(((2b^3 −9abc+27a^2 d−(√((2b^3 −9abc+27a^2 d)^2 −4(b^2 −3ac)^3 )))/2))^(1/3) x_2 =−(b/(3a))−((1−i(√3))/(6a))((((2b^3 −9abc+27a^2 d+(√((2b^3 −9abc+27a^2 d)^2 −4(b^2 −3ac)^3 )))/2)+))^(1/3) ((1+i(√3))/(6a))(((2b^3 −9abc+27a^2 d−(√((2b^3 −9abc+27a^2 d)^2 −4(b^2 −3ac)^3 )))/2))^(1/3)

$${x}_{\mathrm{1}} =−\frac{{b}}{\mathrm{3}{a}}−\frac{\mathrm{1}}{\mathrm{3}{a}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}+\sqrt{\left(\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}\right)^{\mathrm{2}} −\mathrm{4}\left({b}^{\mathrm{2}} −\mathrm{3}{ac}\right)^{\mathrm{3}} }}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{3}{a}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}−\sqrt{\left(\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}\right)^{\mathrm{2}} −\mathrm{4}\left({b}^{\mathrm{2}} −\mathrm{3}{ac}\right)^{\mathrm{3}} }}{\mathrm{2}}} \\ $$$${x}_{\mathrm{2}} =−\frac{{b}}{\mathrm{3}{a}}+\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{6}{a}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}+\sqrt{\left(\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}\right)^{\mathrm{2}} −\mathrm{4}\left({b}^{\mathrm{2}} −\mathrm{3}{ac}\right)^{\mathrm{3}} }}{\mathrm{2}}}−\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{6}{a}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}−\sqrt{\left(\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}\right)^{\mathrm{2}} −\mathrm{4}\left({b}^{\mathrm{2}} −\mathrm{3}{ac}\right)^{\mathrm{3}} }}{\mathrm{2}}} \\ $$$${x}_{\mathrm{2}} =−\frac{{b}}{\mathrm{3}{a}}−\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{6}{a}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}+\sqrt{\left(\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}\right)^{\mathrm{2}} −\mathrm{4}\left({b}^{\mathrm{2}} −\mathrm{3}{ac}\right)^{\mathrm{3}} }}{\mathrm{2}}+}\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{6}{a}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}−\sqrt{\left(\mathrm{2}{b}^{\mathrm{3}} −\mathrm{9}{abc}+\mathrm{27}{a}^{\mathrm{2}} {d}\right)^{\mathrm{2}} −\mathrm{4}\left({b}^{\mathrm{2}} −\mathrm{3}{ac}\right)^{\mathrm{3}} }}{\mathrm{2}}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com