Factorisation of cubic equation

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

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 $(x+0.682327803)(x-(0.341164+1.16154i))(x-(0.341164-1.16451i))$ .

Abdur Rehman Zahid - 6 years, 5 months ago

niceone

rowegie lambojon - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$