To prove that \( n^3 + 106n^2 = \Theta(n^3) \), we need to show that there exist positive constants \( c_1 \), \( c_2 \), and \( n_0 \) such that for all \( n \geq n_0 \):

\[
c_1 \cdot n^3 \leq n^3 + 106n^2 \leq c_2 \cdot n^3
\]

### Step 1: Analyze the Given Expression

Given:
\[
n^3 + 106n^2
\]

We can split this expression into two parts:
1. \( n^3 \): This is a cubic term and clearly dominates for large \( n \).
2. \( 106n^2 \): This is a quadratic term and grows slower than the cubic term \( n^3 \).

### Step 2: Upper Bound (\( O(n^3) \))

For the upper bound, we want to find a constant \( c_2 \) such that:

\[
n^3 + 106n^2 \leq c_2 \cdot n^3
\]

We can factor out \( n^3 \) from the expression:

\[
n^3 + 106n^2 = n^3(1 + \frac{106}{n})
\]

For large \( n \), the term \( \frac{106}{n} \) becomes very small. Therefore:

\[
1 + \frac{106}{n} \leq 2 \text{ for sufficiently large } n
\]

So:
\[
n^3 + 106n^2 \leq 2n^3
\]

Thus, we can choose \( c_2 = 2 \), and this shows that \( n^3 + 106n^2 = O(n^3) \).

### Step 3: Lower Bound (\( \Omega(n^3) \))

For the lower bound, we need to find a constant \( c_1 \) such that:

\[
c_1 \cdot n^3 \leq n^3 + 106n^2
\]

Since \( n^3 \) is part of the expression \( n^3 + 106n^2 \), it is clear that:

\[
n^3 \leq n^3 + 106n^2
\]

This inequality holds with \( c_1 = 1 \), showing that \( n^3 + 106n^2 = \Omega(n^3) \).

### Conclusion: \( \Theta(n^3) \) Bound

Since we've shown that:

\[
c_1 \cdot n^3 \leq n^3 + 106n^2 \leq c_2 \cdot n^3
\]

for \( c_1 = 1 \) and \( c_2 = 2 \), it follows that:

\[
n^3 + 106n^2 = \Theta(n^3)
\]

This means the function \( n^3 + 106n^2 \) has both an upper and lower bound proportional to \( n^3 \), proving that it grows asymptotically as \( n^3 \).

Would you like more details on any part of this explanation?

Here are 5 related questions you might find interesting:
1. How would you prove that \( n^3 + n^2 + n + 1 = \Theta(n^3) \)?
2. What is the difference between \( \Theta(n^3) \) and \( O(n^3) \)?
3. How do you prove that a function is \( \Omega(n^2) \)?
4. What happens if we add a term like \( \log n \) to \( n^3 + 106n^2 \)?
5. Can \( \Theta \)-notation be applied to non-polynomial functions, like exponential functions?

**Tip:** Remember that \( \Theta(g(n)) \) notation provides both a tight upper and lower bound, meaning the function grows at the same rate as \( g(n) \) within constant factors.

Explore how to prove asymptotic equivalence using Theta notation with an example of polynomial growth: n^3 + 106n^2. Learn about upper and lower bounds in asymptotic analysis and their relevance in algorithm complexity.

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