Find it.

Let the entire expression be equal to $x$ .Then we have that: $x^3=26+15\sqrt{3}+26-15\sqrt{3}+3\left(\sqrt[3]{26-15\sqrt{3}}\right)\left(\sqrt[3]{26+15\sqrt{3}}\right)\left(\sqrt[3]{26-15\sqrt{3}}+\sqrt[3]{26+15\sqrt{3}}\right)$ Since $\sqrt[3]{26-15\sqrt{3}}+\sqrt[3]{26+15\sqrt{3}}=x$ ,so: $x^3=52+3\times\sqrt[3]{676-675}\times x\\ x^3=52+3\sqrt[3]{1}x\\x^3=52+3x\\ x^3-3x-52=0$ So we have reduced the question to finding a root of the equation.In this case, the Rational Root Test reveals that $4$ is a root.However this solution is incomplete if we stop here and don't examine the other 2 roots.Dividing $x^3-3x-52$ by $x-4$ ,we get $x^2+4x+13$ .Upon checking the discriminant: $b^2-4ac=4^2-4(1)(13)=16-52=-36$ As the discriminant is smaller than 0,so the quadratic equation has non-real complex roots,but the expression in the question clearly is a real number.Hence we are left with 1 solution,that is $\boxed{4}$

(26-15 * 3^0.5) = (2-3^0.5)^3; (26-15 * 3^0.5)^(1/3) = (2-3^0.5); 2-(3^0.5)+2+(3^0.5) = 4

The first part of the question is 0.

The second part of the question is 3.713.

The answer is 0 + 3.713 = 3.713

But the answer does not support decimals.

Thus the answer is rounded off to the nearest integer.

So the answer is 4.