Prove that: $n^2+3n^3 + 6^{lgn} is $ $\theta(n^3)$

Question

I'm asked to prove that: $n^2+3n^3 + 6^{lgn} is $ $\theta(n^3)$

I know that for Big O, I need to show: $f(n) <= c*g(n)$ But I'm not sure how to show this, since it involves theta.

Any help would be greatly appreciated.

Answer 1

Let $f(n) = 3n^3+n^2+6^{\log_2 n} = 3n^3+n^2+n^{\log_2 6}$ and $g(n)=n^3$.

Show both $f(n) = O(g(n))$ and $g(n) = O(f(n))$. This is exactly showing $f(n) = \Theta(g(n))$.