Question

(x-3)(x^2+3x+9)-x(x-4)(x+4)=21

Answer 1

Ответ:

Пошаговое объяснение:

(x^3-3^3)-x(x^2-4^2)=21

x^3-27-x^3+16x=21

16x=48

x=3

Answer 2

To solve the equation $(x-3)(x^2+3x+9) - x(x-4)(x+4) = 21$, you can start by simplifying the expression on both sides of the equation:

$(x-3)(x^2+3x+9) - x(x-4)(x+4) = 21$

Now, let's expand and simplify each term:

$(x-3)(x^2+3x+9) = x^3 + 3x^2 + 9x - 3x^2 - 9x - 27$

$- x(x-4)(x+4) = -x(x^2 - 16) = -x^3 + 16x$

Now, the equation becomes:

$(x^3 + 3x^2 + 9x - 3x^2 - 9x - 27) - (-x^3 + 16x) = 21$

Next, simplify the equation:

$x^3 + 3x^2 + 9x - 3x^2 - 9x - 27 + x^3 - 16x = 21$

Now, combine like terms:

$x^3 + x^3 + 3x^2 - 3x^2 + 9x - 9x - 16x - 27 = 21$

Simplify further:

$2x^3 - 16x - 27 = 21$

Now, subtract 21 from both sides of the equation:

$2x^3 - 16x - 27 - 21 = 0$

$2x^3 - 16x - 48 = 0$

Now, you have a cubic equation. To solve for x, you can factor out a common factor of 2:

$2(x^3 - 8x - 24) = 0$

Now, you can try to factor the cubic polynomial inside the parentheses. One root is obvious: x = 4, because when you plug that into the polynomial, you get:

$4^3 - 8(4) - 24 = 64 - 32 - 24 = 8 - 24 = -16$

So, one solution is x = 4. Now, let's find the other solutions. We can use synthetic division to find the remaining roots:

Dividing $x^3 - 8x - 24$ by $(x - 4)$, we get:

$$\begin{array}{c|ccc} 4 & 1 & 0 & -8 & -24 \\ \hline & & 4 & 16 & 32 \\ \end{array}$$

The remainder is 32. So, the polynomial $x^3 - 8x - 24$ can be factored as:

$(x - 4)(x^2 + 4x + 8)$

Now, we need to solve the quadratic equation $x^2 + 4x + 8 = 0$. You can use the quadratic formula to find the roots:

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

In this case, $a = 1$, $b = 4$, and $c = 8$. Plugging these values into the formula:

$$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(8)}}{2(1)}$$