Find the real cube root of
`99-70sqrt(2)`.

Answer

To find the real cube root of \( 99 - 70\sqrt{2} \), we can express the term in a form that allows us to use the identity for the cube of a binomial.

### Step-by-Step Solution:

1. **Identify the Expression**: We need to find \( \sqrt[3]{99 - 70\sqrt{2}} \).

2. **Express as a Cube**: We want to express \( 99 - 70\sqrt{2} \) in the form \( (a - b)^3 \). The expansion of \( (a - b)^3 \) is given by:
\[ a^3 - 3a^2b + 3ab^2 - b^3 \]

3. **Break Down the Expression**: We can try to find suitable values for \( a \) and \( b \) such that:
\[ a^3 - b^3 = 99 \quad \text{and} \quad -3a^2b + 3ab^2 = -70\sqrt{2} \]

4. **Choose Values for \( a \) and \( b \)**: Let's assume \( a = 3 \) and \( b = 2\sqrt{2} \). We calculate:
- \( a^3 = 3^3 = 27 \)
- \( b^3 = (2\sqrt{2})^3 = 8 \cdot 2\sqrt{2} = 16\sqrt{2} \)

5. **Calculate \( 99 \)**: We need to check if:
\[ a^3 - b^3 = 27 - 16\sqrt{2} \]
We need to adjust our values. Let's try \( a = 9 \) and \( b = 2\sqrt{2} \):
- \( a^3 = 9^3 = 729 \)
- \( b^3 = (2\sqrt{2})^3 = 16\sqrt{2} \)

6. **Recalculate**: We can also try \( a = 3 \) and \( b = 2\sqrt{2} \) again:
- \( 3^3 = 27 \)
- \( 2\sqrt{2}^3 = 16\sqrt{2} \)

7. **Final Values**: After some trials, we find:
\[ a = 3, \quad b = 2\sqrt{2} \]
Thus:
\[ (3 - 2\sqrt{2})^3 = 99 - 70\sqrt{2} \]

8. **Conclusion**: Therefore, the real cube root of \( 99 - 70\sqrt{2} \) is:
\[ \sqrt[3]{99 - 70\sqrt{2}} = 3 - 2\sqrt{2} \]

### Final Answer:
\[ \sqrt[3]{99 - 70\sqrt{2}} = 3 - 2\sqrt{2} \]