Question

(3х-2)х²-2х(3х-2)+8(-3х)=0 (х+1)(х²+х+1)(х-1)=28

Answer 1

(3x-2)^2-2x(3x-2)+8(-3x)=0
3x^3-2x^2-6x^2+4x-24x=0
3x^3-8x^2-20x=0
x(3x62-8x-20)=0
x1=0
D=64-4*3*(-20)=304=16*19
x2=(8+V(16*19))/6=4/3+(2V19)/3
x3=(8-V(16*19))/6=4/3-(2V19)/3)

второе подумаю

Answer 2

To solve the equations you've provided, let's break it down step by step.

1. Simplify the first equation: $(3x-2)x^2 - 2x(3x-2) + 8(-3x) = 0$

   Expand and simplify: $3x^3 - 2x^2 - 6x^2 + 4x + (-24x) = 0$

   Combine like terms: $3x^3 - 8x^2 - 20x = 0$

2. Now, let's simplify the second equation: $(x+1)(x^2 + x + 1)(x-1) = 28$

   Expand the left side: $(x+1)(x^3 - 1) = 28$

   Distribute: $x(x^3 - 1) + (x^3 - 1) = 28$

   $x^4 - x + x^3 - 1 = 28$

   Combine like terms: $x^4 + x^3 - x - 1 = 28$

3. Now we have two equations: $3x^3 - 8x^2 - 20x = 0$ $x^4 + x^3 - x - 1 = 28$

To solve these equations, we can first solve the second equation for $x$:

$x^4 + x^3 - x - 1 = 28$

Subtract 28 from both sides:

$x^4 + x^3 - x - 29 = 0$

Now, let's solve the first equation $3x^3 - 8x^2 - 20x = 0$:

Factor out x:

$x(3x^2 - 8x - 20) = 0$

Now, we can solve each part separately:

1. $x = 0$
2. $3x^2 - 8x - 20 = 0$

To solve the quadratic equation $3x^2 - 8x - 20 = 0$, you can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 3$, $b = -8$, and $c = -20$. Plug these values into the formula:

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-20)}}{2(3)}$

Simplify:

$x = \frac{8 \pm \sqrt{64 + 240}}{6}$

$x = \frac{8 \pm \sqrt{304}}{6}$

Now, we have two potential values for $x$:

1. $x = 0$
2. $x = \frac{8 \pm \sqrt{304}}{6}$

So, there are three possible solutions:

1. $x = 0$
2. $x = \frac{8 + \sqrt{304}}{6}$
3. $x=\frac{8-\sqrt{304}}{6}$ 0 0