a crazy equation

The answer should be $\sqrt[3]{8}+ \sqrt{21}+ \sqrt[3]{8}-\sqrt{21} =2+2=\boxed{4}$

The problem is incorrect in my opinion. This needs correct LaTex formatting to be clear and the question should be $\sqrt[3]{{8}+3\sqrt{21}}+\sqrt[3]{{8}-3\sqrt{21}}$ to give answer $\boxed{1}$ .

Let’s write:

a = 8 + 3√21 b = 8 – 3√21

So then:

x = ∛a + ∛b

Notice that the cube roots of a and b are real numbers (since every real number has a real cube root), so the value of x we want is also a real number. To simplify, let’s cube both sides:

x3 = (∛a + ∛b)3

x3 = a + b + 3(∛a)2(∛b) + 3(∛b)2(∛a)

We can then write two copies of each squared term, and then group the cube roots, to get:

x3 = a + b + 3∛(a(ab)) + 3∛(b(ba))

Now we utilize that a and b are square root conjugates, so we can simplify to a whole number:

ab = ba = (8 + 3√21)(8 – 3√21) = 64 – 9(21) = -125 = (-5)3

Substituting back and simplifying gives:

x3 = a + b + 3∛(a(-5)3) + 3∛(b(-5)3)

x3 = a + b + 3(-5)(∛a) + 3(-5)(∛b)

x3 = a + b – 15(∛a + ∛b)

Now recall x = ∛a + ∛b, which means:

x3 = a + b – 15x

Furthermore a + b = 16, so we get a simple polynomial:

x3 = 16 – 15x x3 + 15x – 16 = 0

We want to find a real value of x. One strategy is to use special values, and we get lucky in this case: x = 1 is a solution.

Consequently, (x – 1) is a factor. We can factor to check for other solutions, and we get:

(x – 1)(x2 + x + 16) = 0, but 0 is not counted (check the problem!)

The quadratic equation can be solved using the quadratic formula, but both of its roots will have non-zero imaginary parts. These are extraneous roots since the original expression was a real number.

Hence, we have the solution:

1 = ∛(8 + 3√21) + ∛(8 – 3√21)