Решите уравнение:
а) $(3x + 1)^3 = 27x^2(x + 1) + 8x + 2$;
б) $4x^2(2x + 9) = (2x + 3)^3 + 12(3x + 1)$.
$(3x + 1)^3 = 27x^2(x + 1) + 8x + 2$
$27x^3 + 3 * (3x)^2 * 1 + 3 * 3x * 1^2 + 1 = 27x^3 + 27x^2 + 8x + 2$
$27x^3 + 27x^2 + 9x + 1 = 27x^3 + 27x^2 + 8x + 2$
$27x^3 + 27x^2 + 9x - 27x^3 - 27x^2 - 8x = 2 - 1$
x = 1
$4x^2(2x + 9) = (2x + 3)^3 + 12(3x + 1)$
$8x^3 + 36x^2 = 8x^3 + 3 * (2x)^2 * 3 + 3 * 2x * 3^2 + 27 + 36x + 12$
$8x^3 + 36x^2 = 8x^3 + 36x^2 + 54x + 27 + 36x + 12$
$8x^3 + 36x^2 - 8x^3 - 36x^2 - 54x - 36x = 39$
−90x = 39
$x = -\frac{39}{90} = -\frac{13}{30}$
Пожауйста, оцените решение