if x^3 +(1/x^3 )=52, find the value of x^4 +(1/x^4 )=?


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Question Number 119193 by mr W last updated on 22/Oct/20

$i f x^{3} + \frac{1}{x^{3}} = 52, f i n d t h e v a l u e o f$ $x^{4} + \frac{1}{x^{4}} = ?$
Commented by PRITHWISH SEN 2 last updated on 22/Oct/20

$x + \frac{1}{x} = 4, - 2 + 3 i, - 2 - 3 i$ $x^{4} + \frac{1}{x^{4}} = {(x + \frac{1}{x})}^{4} - 4 {(x + \frac{1}{x})}^{2} + 2$ $∴ x^{4} + \frac{1}{x^{4}} = 194 or - 129 + 168 i or - 105 - 168 i$
Commented by mr W last updated on 22/Oct/20

$t h a n k s!$
Commented by PRITHWISH SEN 2 last updated on 22/Oct/20

$welcome$
	Answered by TANMAY PANACEA last updated on 22/Oct/20
	$(x+(1/x))^3 −3(x+(1/x))=52 a^3 −3a−52 =a^3 −3a−(64−12) =a^3 −64−3(a−4) =(a−4)(a^2 +4a+16)−3(a−4) =(a−4)(a^2 +4a+13) =(a−4){(a+2)^2 +9} so a=4→x+(1/x)=4 x^2 +(1/x^2 )=(x+(1/x))^2 −2=14 x^4 +(1/x^4 )=(x^2 +(1/x^2 ))^2 −2=196−2=194$
	${(x + \frac{1}{x})}^{3} - 3 (x + \frac{1}{x}) = 52$ $a^{3} - 3 a - 52$ $= a^{3} - 3 a - (64 - 12)$ $= a^{3} - 64 - 3 (a - 4)$ $= (a - 4) (a^{2} + 4 a + 16) - 3 (a - 4)$ $= (a - 4) (a^{2} + 4 a + 13)$ $= (a - 4) {{(a + 2)}^{2} + 9}$ $s o a = 4 \to x + \frac{1}{x} = 4$ $x^{2} + \frac{1}{x^{2}} = {(x + \frac{1}{x})}^{2} - 2 = 14$ $x^{4} + \frac{1}{x^{4}} = {(x^{2} + \frac{1}{x^{2}})}^{2} - 2 = 196 - 2 = 194$
	Commented by mr W last updated on 22/Oct/20

	$c o r r e c t, t h a n k s!$
	Commented by TANMAY PANACEA last updated on 22/Oct/20

	$s i r i h a v e n o t c o n s i d e r e d c o m l e x s o l u t i o n$
	Answered by 1549442205PVT last updated on 23/Oct/20

	$x^{3} + \frac{1}{x^{3}} = 52 \Leftrightarrow x^{6} - {52 x}^{3} + 1 = 0$ $Δ^{'} = 26^{2} - 1 = 675 = {(15 \sqrt{3})}^{2}$ $\Rightarrow x^{3} = 26 \pm 15 \sqrt{3} = {(2 \pm \sqrt{3})}^{3} \Rightarrow x = 2 \pm \sqrt{3}$ $Since (2 + \sqrt{3}) (2 - \sqrt{3}) = 1 \Rightarrow x + \frac{1}{x}$ $= 2 + \sqrt{3} + 2 - \sqrt{3} = 4$ $\Rightarrow (x^{4} + \frac{1}{x^{4}}) = (x^{2} + \frac{1}{x^{2}}) - 2 = {[{(x + \frac{1}{x})}^{2} - 2]}^{2} - 2$ $= {(4^{2} - 2)}^{2} - 2 = 14^{2} - 2 = 194$
	Answered by malwan last updated on 24/Oct/20

	$x^{6} - 52 x^{3} + 1 = 0$ $t h i s e q u a t i o n h a s 6 s o l u t i o n s$ $4 c o m p l e x$ $a n d 2 r e a l x = 2 \pm \sqrt{3}$ $x = 2 + \sqrt{3}$ $\Rightarrow x^{4} + \frac{1}{x^{4}} = (97 + 56 \sqrt{3}) + (\frac{1}{97 + 56 \sqrt{3}})$ $= (97 + 56 \sqrt{3}) + (\frac{97 - 56 \sqrt{3}}{9409 - 9408})$ $= 97 + 97 = 194$ $a n d t h e s a m e w a y$ $x = 2 - \sqrt{3} \Rightarrow x^{4} + \frac{1}{x^{4}} = 194$
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