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Question Number 61140 by MJS last updated on 29/May/19

can we find an exact solution?  t^6 +4t^4 −12t^3 +24t^2 −24t+8=0

$$\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution}? \\ $$$${t}^{\mathrm{6}} +\mathrm{4}{t}^{\mathrm{4}} −\mathrm{12}{t}^{\mathrm{3}} +\mathrm{24}{t}^{\mathrm{2}} −\mathrm{24}{t}+\mathrm{8}=\mathrm{0} \\ $$

Commented by ajfour last updated on 04/Jun/19

whats the source of this question,  Is there a way to solve it, afterall  Sir?

$${whats}\:{the}\:{source}\:{of}\:{this}\:{question}, \\ $$$${Is}\:{there}\:{a}\:{way}\:{to}\:{solve}\:{it},\:{afterall} \\ $$$${Sir}? \\ $$

Answered by ajfour last updated on 30/May/19

{t^2 (t^2 +2)^2 +20t^2 +8}^2 =144t^2 (t^2 +2)^2   let t^2 =s  ⇒ {s(s+2)^2 +20s+8}^2 =144s(s+2)^2   ......

$$\left\{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} +\mathrm{20}{t}^{\mathrm{2}} +\mathrm{8}\right\}^{\mathrm{2}} =\mathrm{144}{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} \\ $$$${let}\:{t}^{\mathrm{2}} ={s} \\ $$$$\Rightarrow\:\left\{{s}\left({s}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{20}{s}+\mathrm{8}\right\}^{\mathrm{2}} =\mathrm{144}{s}\left({s}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$...... \\ $$

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