(x+1)^3 =1 x=?


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Question Number 195017 by mathlove last updated on 22/Jul/23

${(x + 1)}^{3} = 1 x = ?$
	Answered by Rasheed.Sindhi last updated on 22/Jul/23

	${(x + 1)}^{3} = 1 x = ?$ $L e t x + 1 = y$ $y^{3} - 1 = 0$ $(y - 1) (y^{2} + y + 1) = 0$ $y - 1 = 0 ∣ y^{2} + y + 1 = 0$ $(x + 1) - 1 = 0 ∣ {(x + 1)}^{2} + (x + 1) + 1 = 0$ $x = 0 ✓ ∣ x^{2} + 2 x + 1 + x + 1 + 1 = 0$ $∣ x^{2} + 3 x + 3 = 0$ $x = \frac{- 3 \pm \sqrt{9 - 12}}{2}$ $x = \frac{- 3 \pm i \sqrt{3}}{2} ✓$
	Commented by mathlove last updated on 22/Jul/23

	$t h a n k s s i r$
	Answered by Rasheed.Sindhi last updated on 22/Jul/23

	$AnOther way$ $x^{3} + 3 x^{2} + 3 x + 1 = 1$ $x^{3} + 3 x^{2} + 3 x = 0$ $x (x^{2} + 3 x + 3) = 0$ $x = 0 ∣ x = \frac{- 3 \pm \sqrt{9 - 12}}{2}$ $∣ x = \frac{- 3 \pm i \sqrt{3}}{2}$
	Answered by Rasheed.Sindhi last updated on 22/Jul/23

	$S t i l l a n o t h e r w a y$ ${(x + 1)}^{3} = 1 x = ?$ $x + 1 = \sqrt[3]{1}$ $x + 1 = 1, ω, ω^{2}$ $x = 0, ω - 1, ω^{2} - 1$ $x = \frac{- 1 \pm i \sqrt{3}}{2} - 1 = \frac{- 3 \pm i \sqrt{3}}{2}$
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