Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 158883 by akolade last updated on 10/Nov/21

Commented by Rasheed.Sindhi last updated on 10/Nov/21

$${\left(\underset{a}{\underbrace{\sqrt{3}+x}}\right)}^{12}+{\left(\underset{b}{\underbrace{\sqrt{3}-x}}\right)}^{12}=172928;x=?$$$$a+b=2\sqrt{3};ab=3-x^2=y\left(say\right)$$$$\bullet a^3+b^3={(a+b)}^3-3ab(a+b)$$$$={\left(2\sqrt{3}\right)}^3-3y\left(2\sqrt{3}\right)$$$$=2\sqrt{3}(12-3y)=6\sqrt{3}(4-y)$$$$\bullet a^6+b^6={(a^3+b^3)}^2-2{\left(ab\right)}^3$$$$={\left(6\sqrt{3}(4-y)\right)}^2-2y^3$$$$=108(16-8y+y^2)-2y^3$$$$=1728-864y+108y^2-2y^3$$$$\bullet a^{12}+b^{12}={(a^6+b^6)}^2-2{\left(ab\right)}^6$$$$={(1728-864y+108y^2-2y^3)}^2-2y^6$$$$\bullet {(1728-864y+108y^2-2y^3)}^2-2y^6$$$$=172928$$$$Withthehelpofadvancedcalculator$$$$y=2\Rightarrow 3-x^2=2\Rightarrow x=\pm 1$$

Commented by 1549442205PVT last updated on 11/Nov/21

$$ThankyouSirSindhiforyourshortest$$$$wayofcalculation$$

Commented by Rasheed.Sindhi last updated on 11/Nov/21

$$\mathbb{Y}{\mathrm{ou}}^{\prime }\mathrm{re}\mathbb{W}\mathrm{elcome}\mathrm{sir}!$$$$\mathbb{Y}{\mathrm{ou}}^{\prime }\mathrm{re}\mathrm{also}\mathrm{an}\mathrm{old}\mathrm{forum}-\mathrm{friend}.\mathbb{I}$$$$\mathrm{remember}\mathrm{your}\mathrm{old}\mathrm{valueable}\mathrm{posts}.$$

Commented by akolade last updated on 12/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by Rasheed.Sindhi last updated on 10/Nov/21

$${\left(\underset{a}{\underbrace{\sqrt{3}+x}}\right)}^{12}+{\left(\underset{b}{\underbrace{\sqrt{3}-x}}\right)}^{12}=172928;x=?$$$$\mathrm{Simplifying}\mathrm{LHS}:$$$$a+b=\sqrt{3}+x+\sqrt{3}-x=2\sqrt{3}$$$$ab={\left(\sqrt{3}\right)}^2-{\left(x\right)}^2=3-x^2$$$$a^3+b^3={(a+b)}^3-3ab(a+b)$$$$={\left(2\sqrt{3}\right)}^3-3(3-x^2)\left(2\sqrt{3}\right)$$$$=2\sqrt{3}(12-9+3x^2)=2\sqrt{3}(3+3x^2)$$$$=6\sqrt{3}(1+x^2)$$$$a^6+b^6={(a^3+b^3)}^2-2{\left(ab\right)}^3$$$$={\left(6\sqrt{3}(1+x^2)\right)}^2-2{(3-x^2)}^3$$$$=108{(1+x^2)}^2-2{(3-x^2)}^3$$$$=2x^6+90x^4+270x^2+54$$$$⋮$$$$a^{12}+b^{12}={(a^6+b^6)}^2-2{\left(ab\right)}^6$$$$⋮$$$$\mathrm{Solving}\mathrm{the}\mathrm{equation}:$$$$▸2x^{12}+396x^{10}+8910x^8+49896x^6$$$$+80190x^4+32076x^2+1458$$$$=172928$$$$▸2x^{12}+396x^{10}+8910x^8+49896x^6$$$$+80190x^4+32076x^2-171470=0$$$$▸x^{12}+198x^{10}+4455x^8+24948x^6$$$$+40095x^4+16038x^2-85735=0$$$$Withthehelpofcalculator$$$$x=\pm 1$$

Commented by 1549442205PVT last updated on 11/Nov/21

$$x^{12}+198x^{10}+4455x^8+24948x^6+40095x^4+16038x^2-85735=0$$$$\iff (x^2-1)(x^{10}+199x^8+4654x^6+29602x^4+69697x^2+85735)=0$$$$\iff x^2=1!Asthisisbettter,SirShindhi$$

Commented by Rasheed.Sindhi last updated on 11/Nov/21

$$\mathcal{T}han\mathcal{X}sirPVTforcomplement.$$$$\mathcal{T}hisisevenimportantformebecause$$$$I{don}^{\prime }texpectanyfeedbackfromthe$$$$questioner.$$$$Youraproachwasalsogoodalthough$$$$somewhatmorelabarious.$$$$\left(sirmynameisSindhinotShindhi\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com