x+y=1 x^5 +xy+y^5 =x^2 y^(2 ) faind the volue of x^(2022) +xy+y^(2022) =?


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 174029 by mathlove last updated on 23/Jul/22

$x + y = 1$ $x^{5} + x y + y^{5} = x^{2} y^{2}$ $f a i n d t h e v o l u e o f x^{2022} + x y + y^{2022} = ?$
Commented by MJS_new last updated on 23/Jul/22

$y = 1 - x$ $4 x^{4} - 8 x^{3} + 8 x^{2} - 4 x + 1 = 0$ ${(2 x^{2} - 2 x + 1)}^{2} = 0$ $x = \frac{1}{2} \pm \frac{1}{2} i \land y = conj (x) \Rightarrow x^{n} + y^{n} = 0 \land x y = \frac{1}{2}$ $answer is \frac{1}{2}$
	Answered by mr W last updated on 23/Jul/22

	$x^{2} + y^{2} + 2 x y = 1$ $x^{2} + y^{2} = 1 - 2 x y$ $(x^{2} + y^{2}) (x + y) = 1 - 2 x y$ $x^{3} + y^{3} + x y (x + y) = 1 - 2 x y$ $x^{3} + y^{3} = 1 - 3 x y$ $x^{4} + y^{4} + 2 {(x y)}^{2} = 1 - 4 x y + 4 {(x y)}^{2}$ $x^{4} + y^{4} = 1 - 4 x y + 2 {(x y)}^{2}$ $(x^{4} + y^{4}) (x + y) = 1 - 4 x y + 2 {(x y)}^{2}$ $x^{5} + y^{5} + x y (x^{3} + y^{3}) = 1 - 4 x y + 2 {(x y)}^{2}$ $x^{5} + y^{5} + x y (1 - 3 x y) = 1 - 4 x y + 2 {(x y)}^{2}$ $x^{5} + y^{5} = 1 - 5 x y + 5 {(x y)}^{2}$ ${(x y)}^{2} - x y = 1 - 5 x y + 5 {(x y)}^{2}$ $0 = 1 - 4 x y + 4 {(x y)}^{2}$ ${(2 x y - 1)}^{2} = 0$ $\Rightarrow x y = \frac{1}{2}$ $x, y a r e r o o t s o f$ $z^{2} - z + \frac{1}{2} = 0$ $z = \frac{1}{2} (1 \pm i) = \frac{1}{\sqrt{2}} (\frac{1}{\sqrt{2}} \pm \frac{i}{\sqrt{2}})$ $= \frac{1}{\sqrt{2}} [\cos (\pm \frac{π}{4}) + i \sin (\pm \frac{π}{4})]$ $z_{1}^{2022} = \frac{1}{2^{1011}} [\cos (\frac{2022 π}{4}) + i \sin (\frac{2022 π}{4})]$ $z_{1}^{2022} = \frac{1}{2^{1011}} [\cos (505 π + \frac{π}{2}) + i \sin (505 π + \frac{π}{2})]$ $z_{1}^{2022} = - \frac{i}{2^{1011}}$ $z_{2}^{2022} = \frac{1}{2^{1011}} [\cos (- \frac{2022 π}{4}) + i \sin (- \frac{2022 π}{4})]$ $z_{2}^{2022} = \frac{1}{2^{1011}} [\cos (- 505 π - \frac{π}{2}) + i \sin (- 505 π - \frac{π}{2})]$ $z_{2}^{2022} = \frac{i}{2^{1011}}$ $\Rightarrow x^{2022} + y^{2022} = z_{1}^{2022} + z_{2}^{2022} = 0$ $\Rightarrow x^{2022} + x y + y^{2022} = \frac{1}{2} ✓$
	Commented by behi834171 last updated on 23/Jul/22

	$g r e a t s i r m r W .$
	Commented by Tawa11 last updated on 23/Jul/22

	$Great sir$
	Commented by mathlove last updated on 23/Jul/22

	$t h a n k s m r W$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum