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Factorisation of cubic equation

can we factories $$x^3+x+1=0$$

Note by Chinmay Sangawadekar
3 years, 6 months ago

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Compare it with the identity $$(u+v)^3-3uv(u+v)-(u^3+v^3)=0$$.We get the following system of equations: $u+v=x\\-3uv=1\rightarrow uv=\frac{-1}{3}\rightarrow u^3v^3=\frac{-1}{27}\\u^3+v^3=-1$ Construct an equation in variable $$z$$ having roots $$u^3$$ and $$v^3$$. $(z-u^3)(z-v^3)=z^2-(u^3+v^3)z+u^3v^3=z^2+1z-\frac{1}{27}=0\rightarrow 27z^2+27z-1=0$ Plugging the values in the quadratic formula,we get: $\frac{-27\pm\sqrt{27^2-4(27)(-1)}}{2(27)}\\=\frac{-27\pm\sqrt{729+108}}{54}\\=\frac{-27\pm\sqrt{837}}{54}\\=\frac{-27\pm3\sqrt{93}}{54}=\frac{-9\pm\sqrt{93}}{18}$ So $$u^3=\frac{-9+\sqrt{93}}{18}\rightarrow u=\sqrt[3]{\frac{-9+\sqrt{93}}{18}}\approx 0.329452338$$and $$v^3=\frac{-9-\sqrt{93}}{18}\rightarrow v=\sqrt[3]{\frac{-9-\sqrt{93}}{18}}\approx -1.011780141$$and $$x=u+v=0.329452338-1.011780141=-0.682327803$$.Dividing by $$(x+0.682327803)$$ and discarding the remainder (since we are working with approximations there will be some very small remainders which would be 0 had we divided by the exact value) we get $$x^2-0.682328x+1.46557$$ which has roots $$0.341164\pm1.16154i$$.So the cubic can be factored as $$(x+0.682327803)(x-(0.341164+1.16154i))(x-(0.341164-1.16451i))$$.

- 3 years ago

niceone

- 3 years, 6 months ago