Question

6(y+1)^2+2(y-1)(y^2+y+1)-2(y+1)^3=32

Answer 1

Ответ:

5

Объяснение:

6(y+1)^2+2(y-1)(y^2+y+1)-2(y+1)^3=32

6(y+1)²+2(y³-1)-2(y+1)³=32

6(y²+2y+1)+2y³-2-2(y³+3y²+3y+1)=32

6y²+12y+6+2y³-2-2y³-6y²-6y-2=32

6y+2=32

6y+2-32=0

6y-30=0

6y=30

y=5

Answer 2

To solve the equation $6(y+1)^2+2(y-1)(y^2+y+1)-2(y+1)^3=32$, we'll need to simplify the left-hand side and then solve for the value(s) of $y$ that satisfy the equation.

Let's break down the steps:

1. Expand the expressions:

   Expand each term in the equation to simplify it.

   $$\begin{align*} &6(y+1)^2 + 2(y-1)(y^2+y+1) - 2(y+1)^3 = 32 \\ &6(y^2 + 2y + 1) + 2(y^3 - y^2 + y^2 + y - y + 1) - 2(y^3 + 3y^2 + 3y + 1) = 32 \\ &6y^2 + 12y + 6 + 2y^3 + 2y - 2y + 2 - 2y^3 - 6y^2 - 6y - 2 = 32 \\ &6y^2 + 12y + 6 + 2y + 2 - 2 = 32 \\ &6y^2 + 12y + 6 + 2y = 32 \\ &2y^2 + 14y + 6 = 32 \\ &2y^2 + 14y - 26 = 0 \\ \end{align*}$$

2. Simplify the equation:

   The equation becomes a quadratic equation in terms of $y$:

   $$2y^2 + 14y - 26 = 0$$

3. Solve the quadratic equation:

   To solve the quadratic equation, we can use the quadratic formula:

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

   In this case, $a = 2$, $b = 14$, and $c = -26$.

   $$y = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 2 \cdot (-26)}}{2 \cdot 2}$$

   Calculating the values inside the square root:

   $$y = \frac{-14 \pm \sqrt{196 + 208}}{4}$$
   $$y = \frac{-14 \pm \sqrt{404}}{4}$$
   $$y = \frac{-14 \pm 2\sqrt{101}}{4}$$

   Simplifying further:

   $$y = \frac{-7 \pm \sqrt{101}}{2}$$

So, the solutions for $y$ are:

$$y = \frac{-7 + \sqrt{101}}{2} \quad \text{and} \quad y = \frac{-7 - \sqrt{101}}{2}$$

These are the solutions that satisfy the original equation.