[tex]\sqrt[3]{9+\sqrt{80} } +\sqrt[3]{9-\sqrt{80} } =[/tex]

Предыдущий
вопрос Следующий
вопрос

Let's simplify each cube root separately:

First, let's factor 80 under the square root:

[tex]\sqrt{80} = \sqrt{16 \times 5} = \sqrt{16} \times \sqrt{5} = 4\sqrt{5}[/tex]

So, [tex]\sqrt[3]{9+\sqrt{80}} = \sqrt[3]{9+4\sqrt{5}}[/tex]

Now let's assume that [tex]\sqrt[3]{9+4\sqrt{5}} = a + b[/tex], where a and b are real numbers.

Then, tex^3 = a^3 + 3a^2b + 3ab^2 + b^3 = 9 + 4\sqrt{5}[/tex]

Equating real and imaginary parts, we get:

Since the first equation states that the sum of two cube roots is 9, and when we cube this sum, we get 9 + 4√5.

The solution to this system of equations is a = 1 and b = 2.

Thus, [tex]\sqrt[3]{9+\sqrt{80}} = 1 + 2 = 3[/tex]

Similarly, [tex]\sqrt[3]{9-\sqrt{80}} = \sqrt[3]{9-4\sqrt{5}}[/tex]

The same process as above can be followed and it can be determined that [tex]\sqrt[3]{9-\sqrt{80}} = 1 - 2 = -1[/tex]

By adding these two results together, we get:

[tex]\sqrt[3]{9+\sqrt{80}} + \sqrt[3]{9-\sqrt{80}} = 3 + (-1) = 2[/tex]