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Question Number 132494 by mnjuly1970 last updated on 14/Feb/21

Commented by MJS_new last updated on 14/Feb/21

$$\mathrm{sharing}\mathrm{these}\mathrm{transformations}$$$$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{check}\mathrm{all}\mathrm{solutions}!!!$$$$$$$$a^{1/2}+b^{1/2}=c^{1/2}$$$${\mid }^2\mathrm{and}\mathrm{transforming}$$$$2a^{1/2}{b}^{1/2}=-a-b+c$$$${\mid }^2$$$$4ab={(a+b-c)}^2$$$$$$$$a^{1/3}+b^{1/3}=c^{1/3}$$$${\mid }^3\mathrm{and}\mathrm{transforming}$$$$3a^{1/3}{b}^{1/3}(a^{1/3}+b^{1/3})=-a-b+c$$$$\mathrm{but}{a}^{1/3}+b^{1/3}=c^{1/3}$$$$3a^{1/3}{b}^{1/3}{c}^{1/3}=-a-b+c$$$${\mid }^3$$$$27abc=-{(a+b-c)}^3$$$$$$$$a^{1/4}+b^{1/4}=c^{1/4}$$$$\mathrm{starting}\mathrm{with}\mathrm{result}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{root}$$$$\mathrm{equation}\mathrm{sbove}$$$$4\mathrm{ab}={(\mathrm{a}+\mathrm{b}-\mathrm{c})}^2\iff {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2-2(\mathrm{ab}+\mathrm{ac}+\mathrm{bc})=0$$$$\Rightarrow$$$$a+b+c-2(a^{1/2}{b}^{1/2}+a^{1/2}{c}^{1/2}+b^{1/2}{c}^{1/2})=0$$$$\mathrm{transforming}$$$$a+b+c-2a^{1/2}{b}^{1/2}=2c^{1/2}(a^{1/2}+b^{1/2})$$$${\mid }^2\mathrm{and}\mathrm{transforming}$$$$a^2+b^2+c^2-6ab-2ac-2bc=4a^{1/2}{b}^{1/2}(a+b+3c)$$$${\mid }^2\mathrm{and}\mathrm{transforming}$$$$8ab(a^2+b^2+17c^2+14ac+14bc)={(a+b-c)}^4$$$$$$$$\mathrm{I}\mathrm{like}\mathrm{these}:$$$$\frac{{(a+b-c)}^2}{4ab}=1$$$$\frac{{(a+b-c)}^3}{27abc}=-1$$$$\frac{{(a+b-c)}^4}{8ab}=a^2+b^2+17c^2+14(a+b)c$$

Commented by MJS_new last updated on 14/Feb/21

$$\mathrm{I}\mathrm{have}\mathrm{these}\mathrm{up}\mathrm{to}{a}^{1/9}+b^{1/9}=c^{1/9}\mathrm{but}\mathrm{I}\mathrm{forgot}$$$$\mathrm{the}\mathrm{paths}...$$

Commented by mnjuly1970 last updated on 14/Feb/21

$$gratefulmrmjs...$$

Answered by MJS_new last updated on 14/Feb/21

$$\mathrm{using}\mathrm{the}{3}^{\mathrm{rd}}\mathrm{above}\mathrm{equation}\mathrm{with}$$$$a=x-1\wedge b=x+1\wedge c=x+3$$$$\mathrm{we}\mathrm{get}$$$$x^4+4x^3+\frac{54}{25}{x}^2-\frac{92}{25}x-\frac{1321}{375}=0$$$$\mathrm{let}x=z-1$$$$z^4-\frac{96}{25}{z}^2-\frac{256}{375}=0$$$$\mathrm{and}\mathrm{this}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{solve}$$$$\mathrm{of}\mathrm{the}4\mathrm{solutions}\mathrm{only}$$$$x=\frac{-15+\sqrt{3(9+4\sqrt{6})}}{15}$$$$\mathrm{solves}\mathrm{the}\mathrm{given}\mathrm{equation}$$

Answered by Tawa11 last updated on 06/Nov/21

$$\mathrm{Great}$$

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