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Question Number 48294 by behi83417@gmail.com last updated on 21/Nov/18

Commented by MJS last updated on 22/Nov/18

$$x^6+x^5+x^4-10x^3+x^2+x+1=0$$$$x=\tan \frac{t}{2}$$$$\mathrm{after}\mathrm{some}\mathrm{transforming}\mathrm{work}\mathrm{we}\mathrm{get}$$$$\frac{{3\sin }^{3}t+{\sin }^{2}t-\sin t-2}{{(1+\cos t)}^3}=0$$$$\sin t=y$$$$3y^3+y^2-y-2=0$$$$y^3+\frac{1}{3}{y}^2-\frac{1}{3}y-\frac{2}{3}=0$$$$y=z-\frac{1}{9}$$$$z^3-\frac{10}{27}z-\frac{457}{729}=0$$$$D=\frac{p^3}{27}+\frac{q^2}{4}=\frac{281}{2916}>0\Rightarrow 1\mathrm{real}\mathrm{solution}$$$$z=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}$$$$z=\frac{1}{9}(\sqrt[3]{\frac{457+27\sqrt{281}}{2}}+\sqrt[3]{\frac{457-27\sqrt{281}}{2}})$$$$y=\frac{1}{9}(\sqrt[3]{\frac{457+27\sqrt{281}}{2}}+\sqrt[3]{\frac{457-27\sqrt{281}}{2}}-1)$$$$t=2n\pi +\arcsin y\vee t=(2n+1)\pi -\arcsin y;n\in \mathbb{Z}$$$$x=\tan \frac{\arcsin y}{2}\vee x=\cot \frac{\arcsin y}{2}$$$$x=\frac{1}{y}\pm \frac{\sqrt{1-y^2}}{y}$$$$...\mathrm{so}\mathrm{we}\mathrm{have}\mathrm{exact}\mathrm{solutions}\mathrm{of}\mathrm{an}\mathrm{unpleasant}\mathrm{shape}$$$$x_1\approx .608053$$$$x_2\approx 1.64459$$

Commented by MJS last updated on 22/Nov/18

$$(x-\frac{1}{y}+\frac{\sqrt{1-y^2}}{y})(x-\frac{1}{y}-\frac{\sqrt{1-y^2}}{y})=x^2-\frac{2}{y}+1$$$$\frac{x^6+x^5+x^4-10x^3+x^2+x+1}{x^2-\frac{2}{y}+1}=$$$$=x^4+(1+\frac{2}{y})x^3+(\frac{2}{y}+\frac{4}{y^2})x^2-(11+\frac{2}{y}-\frac{4}{y^2}-\frac{8}{y^3})x+(1-\frac{24}{y}-\frac{8}{y^2}+\frac{8}{y^3}+\frac{16}{y^4})+F$$$$F=\frac{4((y^2-4)x+2y)}{y^5(x^2-\frac{2}{y}+1)}(3y^2+y^2-y-2)$$$$\mathrm{and}\mathrm{since}F=0\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{same}\mathrm{solution}\mathrm{as}$$$$\mathrm{above}.\mathrm{not}\mathrm{sure}\mathrm{which}\mathrm{method}\mathrm{is}\mathrm{more}\mathrm{comfortable}$$

Commented by behi83417@gmail.com last updated on 22/Nov/18

$$thankyousomuchdearMJS,Ajfour$$$$andsirtanmay.$$

Answered by ajfour last updated on 22/Nov/18

$$let\frac{1}{x}=t$$$$\Rightarrow (1+t)(\frac{1}{t}+t)(\frac{1}{t^2}+t)=12$$$$or(t+1)(t^2+1)(t^3+1)=12t^3$$$$\Rightarrow if\alpha isaroot,then\frac{1}{\alpha }isalso$$$$aroot.$$$$\Rightarrow \mathrm{\Pi }(x-\alpha )(x-\frac{1}{\alpha })=0$$$$or\mathrm{\Pi }[x^2-(\alpha +\frac{1}{\alpha })x+1]=0$$$$let\alpha +\frac{1}{\alpha }=p,\beta +\frac{1}{\beta }=q,\gamma +\frac{1}{\gamma }=r$$$$\mathrm{\Pi }(x^2-px+1)=0$$$$x^6-(p+q+r)x^5+(3+pq+qr+rp)x^4$$$$-(p+q+r+pqr)x^3$$$$+(3+pq+qr+rp)x^2-(p+q+r)x$$$$+1=0$$$$giveneq.(x+1)(x^2+1)(x^3+1)=12x^3$$$$\Rightarrow x^6+x^5+x^4-10x^3+x^2+x+1=0$$$$\Rightarrow p+q+r=-1$$$$pq+qr+rp=-2$$$$pqr=11$$$$p,q,rarethenrootsof$$$$t^3+t^2-2t-11=0$$$$shouldbe{t}^3+t^2-2t-12=0$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\mathbb{BUT}\mathbb{HOW}?$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$t=p\approx 2.19479\left(onlyonerealroot\right).$$$$\Rightarrow \alpha +\frac{1}{\alpha }=p$$$${\alpha }^2-p\alpha +1=0$$$$$$$$someerror...$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Nov/18

$$x^6+x^5+x^4-10x^3+x^2+x+1=0$$$$x^3+x^2+x-10+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}=0$$$$(x^3+\frac{1}{x^3})+(x^2+\frac{1}{x^2})+(x+\frac{1}{x})-10=0$$$$t=x+\frac{1}{x}$$$$t^3-3t+t^2-2+t-10=0$$$$t^3+t^2-2t-12=0$$$$f\left(t\right)=t^3+t^2-2t-12$$$$f\left(0\right)<0$$$$f\left(1\right)<0$$$$f\left(2\right)<0butf\left(3\right)>0$$$$soonerootofeqn{t}^3+t^2-2t-12=0$$$$liesinbetween2and3$$$$nowcalculatingvalueoff(2.2)andf(2.3)using$$$$calculus$$$$f\left(t\right)=t^3+t^2-2t-12$$$$f\left(2\right)=-4$$$$\frac{df}{dt}=\frac{△f}{△t}=3t^2+2t-2$$$$so\frac{df}{dt}=14att=2$$$$\frac{△f}{△t}=\frac{df}{dt}$$$$△f=\frac{df}{dt}△t=14\times 0.2=2.8$$$$sof(2.2)=-4+2.8=-1.2$$$$f(2.3)=-4+4.2=0.2$$$$f(2.2)<0butf(2.3)>0$$$$soonerootofeqn{t}^3+t^2-2t-12=0$$$$liesbetween2.2and2.3$$$$solet2.3>\alpha >2.2issolutionof{t}^3+t^2-2t-12=0$$$$$$$$x+\frac{1}{x}=\alpha$$$$x^2-\alpha x+1=0$$$$x=\frac{\alpha \pm \sqrt{{\alpha }^2-4}}{2}$$$$wecanfind\alpha usingapproxmiatation....$$$$$$$$$$$$$$

Commented by behi83417@gmail.com last updated on 22/Nov/18

$$t^3+t^2-2t-12=0$$$$somtryshowsthatt=2.25isananswer.$$$$x+\frac{1}{x}=2.25\Rightarrow x^2-2.25x+1=0$$$$x=\frac{2.25\pm \sqrt{2.{25}^2-4}}{2}=\frac{2.25\pm 1.032}{2}=$$$$\Rightarrow x=1.64,x=0.61.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Nov/18

$$thankyousir..$$

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