Find the sum of the fifth powers of the roots of x^3 −2x^2 +x−1=0 by applying synthetic division


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Question Number 200465 by deleteduser12 last updated on 19/Nov/23

$Find the sum of the fifth powers of$ $the roots of x^{3} - {2 x}^{2} + x - 1 = 0 by$ $applying synthetic division$
Commented by mr W last updated on 19/Nov/23

$y o u p o s t e d y o u r f i r s t q u e s t i o n o n$ $15 / 09 / 2020 . s i n c e t h e n y o u h a v e n e v e r$ $g i v e n e v e n a s i n g l e o n e f e e d b a c k .$ $e v e n w h e n p e o p l e a r e t a l k i n g t o a$ $s t o n e w a l l, i t g i v e s a n e c h o b a c k . . .$
	Answered by mr W last updated on 19/Nov/23

	$s a y t h e r o o t s a r e a, b, c .$ $a + b + c = 2$ $a b + b c + c a = 1$ $a b c = 1$ ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$ $\Rightarrow a^{2} + b^{2} + c^{2} = 2^{2} - 2 \times 1 = 2$ ${(a b + b c + c a)}^{2} = a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} + 2 a b c (a + b + c)$ $\Rightarrow a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = 1^{2} - 2 \times 1 \times 2 = - 3$ ${(a + b + c)}^{3} = a^{3} + b^{3} + c^{3} - 3 a b c + 3 (a + b + c) (a b + b c + c a)$ $\Rightarrow a^{3} + b^{3} + c^{3} = 2^{3} + 3 \times 1 - 3 \times 2 \times 1 = 5$ $(a^{2} + b^{2} + c^{2}) (a^{3} + b^{3} + c^{3}) = a^{5} + b^{5} + c^{5} + (a + b + c) (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}) - a b c (a b + b c + c a)$ $\Rightarrow a^{5} + b^{5} + c^{5} = 2 \times 5 - 2 \times (- 3) + 1 \times 1 = 17 ✓$
	Answered by ajfour last updated on 19/Nov/23

	$l e t x^{5} = t$ $\Rightarrow x^{2} (2 x^{2} - x + 1) = t$ $2 x (2 x^{2} - x + 1) - (2 x^{2} - x + 1) + x^{2} = t$ $4 (2 x^{2} - x + 1) - 3 x^{2} + 3 x - 1 = t$ $5 x^{2} - x + 3 = t$ $5 (2 x^{2} - x + 1) - x^{2} = (t - 3) x$ $9 (x + t - 3) + 25 = 5 (t + 2) x$ $(5 t + 1) x = 9 t - 2$ $x = \frac{9 t - 2}{5 t + 1}$ ${(9 t - 2)}^{3} - 2 (5 t + 1) {(9 t - 2)}^{2}$ $+ {(5 t + 1)}^{2} (9 t - 2) - {(5 t + 1)}^{3} = 0$ $\underset{i = 1}{\sum^{3}} x_{i}^{5} = \underset{i = 1}{\sum^{3}} t_{i} = - \frac{c o e f f o f t^{2}}{c o e f f o f t^{3}}$ $= - \frac{(- 6 \times 81 + 360 - 216 - 50 + 90 - 75)}{(729 - 810 + 225 - 125)}$ $= - \frac{(- 323)}{19} = 17$
	Commented by mr W last updated on 20/Nov/23

	$g r e a t!$
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