Double Cube Roots

we know that $64000=64 \times {10}^{3}={2}^{6} \times {10}^{3} ={{\left( {{2}^{2}} \right)}^{3}}\times {{10}^{3}}$ so $\sqrt[3]{64000}=\sqrt[3]{{{\left( {{2}^{2}} \right)}^{3}}\times {{10}^{3}}}={2}^{2} \times 10=40$ also we have $68921={41}^{3}$ thus $\sqrt[3]{68921}=\sqrt[3]{{41}^{3}}=41$ hence $\sqrt[2]{40+41}=\sqrt[2]{81}=\sqrt[2]{{9}^{2}}=9$ therefore $\sqrt{9}=\sqrt{{3}^{2}}=3$

68921 (Refer My Note given in the Q)
First check the last three digit-921,. Its last digit is 1 & the last digit of the cube root of 68921 is 1.
68 is greater than 64(cube of 4)
cube root of 68921 is 41.

Similarly, cube root of 64000 is 40.

next square root of 81 (40+41) is 9.

next square root of 9 is 3

The cube root of 64 000 is 40, and since 41 cubed - 40 cubed is 68 921 - 64 000 is 4 921 = 3 × 1640 (4920) + 1, the cube root of 64 000 + 3 × 1640 + 1 is 41. 40 + 41 is 81, and since a square root of a square root is a hypercube root, the 4th root of 81 is 3. Thus, the answer is 3.
Bonus Proof: (n+1)^3 - n^3 = n^3 + 3n^2 + 3n + 1 - n^3 = 3n^2 + 3n +1. Plug in 40 to get 3(40^2 + 40) + 1 =3(1640) + 1.

As written, we know that $(a+b)^ {3} = a^{3} + b^{3}+ 3ab(a+b)$ and just by observation, by looking at the cube root at the right, we can tell that $a = 40$ and $b = 1$ . Then we can simplify that as cube root $(40+1)^{3}$ which of course, is 41.

Looking at the root, we get cube root 64000 which we calculate as 40 + 41, as we just worked out and we square root the answer, which happens to be 81. We square root 81 to make 9, and square root it again to make 3.

Hence the answer is $\boxed{3}$ .

The note is really helpful . But we can't be sure whether a given number is a perfect cube

sqrt(sqrt( cbrt(64000) + cbrt(68921) ))) =sqrt(sqrt(40+41))) =sqrt(sqrt(81))) =sqrt(9) = 3