Question

(x^2+x+1)^2-2x(x+1)=5

Answer 1

(x² + x + 1) * (x² + x +1) - 2x² - 2x = 5

x^4 + x³ + x² + x³ + x² + x + x² + x + 1 - 2x² - 2x = 5

x^4 + 2x³ + x² - 4 = 0

x³(x+2) + (x-2)(x+2) = 0

(x+2)(x³ + x -2) = 0

1) x+2 = 0

   x1= -2

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2) x³ + x -2 = 0

x³ - 1 + x - 1 = 0

(x-1)(x² + x + 1) + (x - 1) = 0

(x-1)(x² + x + 1 + 1) = 0

x - 1= 0

x2 = 1

3) x² + x + 1 + 1 =0

x² + x + 2 = 0

D=b² - 4ac = 1² - 4*1*2 = 1 - 8 = -7

D <0 ⇒ корней нет

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x1 = -2

x2 = 1