Calculate it!

Method 1: Let $x=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\\$ $x-\sqrt[3]{20+14\sqrt{2}}-\sqrt[3]{20-14\sqrt{2}} =0\\$

Using the fact that $a+b+c=0 \implies a^3+b^3+c^3=3abc$ we get $x^3-(20+14\sqrt{2})-(20-14\sqrt{2})=3x(\sqrt[3]{20+14\sqrt{2}})(\sqrt[3]{20-14\sqrt{2}})$ $\implies x^3-40=3x\sqrt[3]{400-392}$ $\implies x^3-40=6x$ $\implies x^3-6x-40=0$

Since from the given options we know that it has one of the roots among $1,2,4,8$ we check these and get that $4$ is a rational root of the above cubic.

$x^3-6x-40=(x-4)(x^2+4x=10$

Since the roots of $x^2+4x+10$ are non real complex and $x$ is the rum of two purely real numbers, only value it can take is $4$ .

So $x=\boxed{4}$ is the answer.

METHOD 2: BASED ON LUCK

Observe that the quantities under the cube roots are the cubes of $2+\sqrt{2}$ and $2-\sqrt{2}$ and we get directly our answer.

But this is my point of view when posing the question and is merely a lucky strike for the solver. So follow method 1

time saving method

1) 14root(2) approx=19.6

2)cuberoot(39.6)+cuberoot(.4)

3)approx (3.something) +(.something) = only one option near this value=4

Get a look! We can see that (20+14sqrt(2)) = (2+sqrt(2))^3 (20- 14sqrt(2)) = (2-sqrt(2))^3 So the expression becomes: 2+sqrt(2)+2-sqrt(2) = 4 P.S : Sorry because I can't use LaTex

The options suggest that the values inside the roots are perfect cubes. Using the formulae of (a+b)^3 and (a-b)^3 we get a^3 + 3ab^2 =20 (1) b^3 + 3ba^2=14root2 (2) Take 'a' common in equation 1 and 'b' common in equation 2. It is clear that 'a' is rational and 'b' is a multiple of root 2. As 'a' is rational, it has to be a factor of 20. It is then easy to see that the value of 'a' is 2('b' comes out as root2). Hence the answer is (a+b)+(a-b) = 4