Question

(a²-3)³-(a²-4)(a²+4)-a²(a4-10a²+ 27)+27=16

Answer 1

(a²-3)³-(a²-4)(a²+4)-a²(a4-10a²+ 27)+27=16

$a^{6}$ + 0 -27 - $4a^{3}$ = 0

$a^{6}$ - $4a^{3}$ - 27 = 0

$t^{2}$ - 4t - 27 = 0

t = 2 + $\sqrt{31}$

t = 2 - $\sqrt{31}$

$a^{3}$ = 2 + $\sqrt{31}$

$a^{3}$ = 2 - $\sqrt{31}$

Ответ: a = $\sqrt[3]{2+\sqrt{31}}$

a = $\sqrt[3]{2-\sqrt{31}}$

Answer 2

Let's simplify the given equation step by step:

(a² - 3)³ - (a² - 4)(a² + 4) - a²(a⁴ - 10a² + 27) + 27 = 16

First, let's expand the cubes:

(a⁶ - 9a⁴ + 27a² - 27) - (a⁴ - 4a² - 4a² - 16) - (a⁶ - 10a⁴ + 27a²) + 27 = 16

Now, combine like terms:

a⁶ - 9a⁴ + 27a² - 27 - a⁴ + 4a² + 4a² + 16 - a⁶ + 10a⁴ - 27a² + 27 = 16

Cancel out like terms:

-9a⁴ + 35a² - 9 = 16

Rearrange the equation:

-9a⁴ + 35a² - 9 - 16 = 0

-9a⁴ + 35a² - 25 = 0

Now, we have a quadratic equation in terms of a². Let's solve it:

Let's substitute x = a²:

-9x² + 35x - 25 = 0

Using the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Where a = -9, b = 35, and c = -25.

x = (-35 ± √(35² - 4(-9)(-25))) / (2(-9))

x = (-35 ± √(1225 - 900)) / (-18)

x = (-35 ± √325) / (-18)

Since we substituted x = a², we need to solve for a:

a² = (-35 ± √325) / (-18)

Taking the square root:

a = ±√((-35 ± √325) / (-18))

So, the solutions for 'a' are:

a = √((-35 + √325) / (-18)) a = √((-35 - √325) / (-18)) a = -√((-35 + √325) / (-18)) a = -√((-35 - √325) / (-18))

Please note that these are the exact solutions to the equation. If you'd like decimal approximations, you can calculate them using a calculator.