Let a be the unique real zero of x^3 +x+1. find the simplest possible way to write ((18)/((a^2 +a+1)^2 )) as polynomial expression in a with ratio coefficients


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Question Number 210310 by Spillover last updated on 05/Aug/24

$L e t a b e t h e u n i q u e r e a l z e r o o f x^{3} + x + 1 .$ $f i n d t h e s i m p l e s t p o s s i b l e w a y t o w r i t e$ $\frac{18}{{(a^{2} + a + 1)}^{2}} a s p o l y n o m i a l e x p r e s s i o n i n a$ $w i t h r a t i o c o e f f i c i e n t s$
Commented by Frix last updated on 07/Aug/24

$Do you know the answer ?$
Commented by AlagaIbile last updated on 07/Aug/24

$6 {(a^{2} + 1)}^{3}$
Commented by Frix last updated on 08/Aug/24

$\frac{18}{{(a^{2} + a + 1)}^{2}} \approx 29.34$ $6 {(a^{2} + 1)}^{3} \approx 18.89$
Commented by Spillover last updated on 08/Aug/24

$14 a^{2} - 10 a + 16$
	Answered by Frix last updated on 08/Aug/24

	$We have a^{3} + a + 1 = 0 \land a \in R$ $\frac{18}{{(a^{2} + a + 1)}^{2}} = \frac{18}{a^{4} + 2 a^{3} + 3 a^{2} + 2 a + 1} =$ $a^{3} = - (a + 1) \Rightarrow a^{4} = - a (a + 1)$ $= \frac{18}{2 a^{3} + 2 a^{2} + a + 1} =$ $a^{3} = - (a + 1)$ $= \frac{18}{2 a^{2} - a - 1} =$ $a^{2} = - \frac{a + 1}{a}$ $= - \frac{18 a}{a^{2} + 3 a + 2} = - \frac{18 a}{(a + 1) (a + 2)} =$ $= \frac{18}{a + 1} - \frac{36}{a + 2}$ $We know a = u^{\frac{1}{3}} + v^{\frac{1}{3}}$ $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $\frac{p}{a + q} = \frac{p}{q + u^{\frac{1}{3}} + v^{\frac{1}{3}}} =$ $= \frac{p}{q + u^{\frac{1}{3}} + v^{\frac{1}{3}}} \times \frac{(q + ω u^{\frac{1}{3}} + ω^{2} v^{\frac{1}{3}}) (q + ω^{2} u^{\frac{1}{3}} + ω v^{\frac{1}{3}})}{(q + ω u^{\frac{1}{3}} + ω^{2} v^{\frac{1}{3}}) (q + ω^{2} u^{\frac{1}{3}} + ω v^{\frac{1}{3}})} =$ $= \frac{p (q^{2} - (u^{\frac{1}{3}} + v^{\frac{1}{3}}) q + u^{\frac{2}{3}} - u^{\frac{1}{3}} v^{\frac{1}{3}} + v^{\frac{2}{3}})}{q^{3} + u + v - 3 u^{\frac{1}{3}} v^{\frac{1}{3}} q} =$ $Cardano gives u, v :$ $a = {(- \frac{1}{2} + \frac{\sqrt{93}}{18})}^{\frac{1}{3}} + {(- \frac{1}{2} - \frac{\sqrt{93}}{18})}^{\frac{1}{3}}$ $u = - \frac{1}{2} + \frac{\sqrt{93}}{18} \land v = - \frac{1}{2} - \frac{\sqrt{93}}{18}$ $u v = - \frac{1}{27} \Rightarrow u + v = - 1 \land u^{\frac{1}{3}} v^{\frac{1}{3}} = - \frac{1}{3}$ $= \frac{p (q^{2} - q a + u^{\frac{2}{3}} + 2 u^{\frac{1}{3}} v^{\frac{1}{3}} + v^{\frac{2}{3}} - 3 u^{\frac{1}{3}} v^{\frac{1}{3}})}{q^{3} + q - 1} =$ $= \frac{p (q^{2} - q a + a^{2} + 1)}{q^{3} + q - 1}$ $\frac{18}{a + 1} - \frac{36}{a + 2} = 18 (x^{2} - x + 2) - 4 (x^{2} - 2 x + 5) =$ $= 2 (7 x^{2} - 5 x + 8)$
	Answered by Spillover last updated on 08/Aug/24

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