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.