a^3 +(1/a^3 )=18 a^4 +(1/a^4 )=?


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Question Number 86825 by M±th+et£s last updated on 31/Mar/20

$a^{3} + \frac{1}{a^{3}} = 18$ $a^{4} + \frac{1}{a^{4}} = ?$
Commented by Ar Brandon last updated on 31/Mar/20

$a^{3} + \frac{1}{a^{3}} = 18 \Rightarrow {(a^{3})}^{2} - 18 a^{3} + 1 = 0$ $a^{3} = \frac{18 \pm \sqrt{{(- 18)}^{2} - 4}}{2} = \frac{18 \pm \sqrt{320}}{2} = 9 \pm 4 \sqrt{5}$ $a = {(9 \pm 4 \sqrt{5})}^{\frac{1}{3}}$ $a^{4} + \frac{1}{a^{4}} = {(9 \pm 4 \sqrt{5})}^{\frac{4}{3}} + \frac{1}{{(9 \pm 4 \sqrt{5})}^{\frac{4}{3}}}$
	Answered by mr W last updated on 31/Mar/20

	${(a + \frac{1}{a})}^{3} = a^{3} + \frac{1}{a^{3}} + 3 (a + \frac{1}{a}) = 18 + 3 (a + \frac{1}{a})$ $l e t A = a + \frac{1}{a}$ $A^{3} - 3 A - 18 = 0$ $(A - 3) (A^{2} + 3 A + 6) = 0$ $\Rightarrow A = 3 = a + \frac{1}{a}$ $a^{4} + \frac{1}{a^{4}} = {(a^{2} + \frac{1}{a^{2}})}^{2} - 2 = {[{(a + \frac{1}{a})}^{2} - 2]}^{2} - 2$ $= {[3^{2} - 2]}^{2} - 2$ $= 47$
	Commented by M±th+et£s last updated on 31/Mar/20

	$t h a n k y o u s i r$
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