Sure! Let's solve the equations step by step.

### 1) \((x+3)(x^2-3x+5)=x^2+3x\)

First, expand the left side:

\[
(x+3)(x^2-3x+5) = x(x^2-3x+5) + 3(x^2-3x+5)
\]

This simplifies to:

\[
x^3 - 3x^2 + 5x + 3x^2 - 9x + 15 = x^3 - 4x + 15
\]

Setting this equal to the right side:

\[
x^3 - 4x + 15 = x^2 + 3x
\]

Rearranging gives:

\[
x^3 - x^2 - 7x + 15 = 0
\]

### 2) \(x^4 - 9x^2 = 4x^2 - 36\)

Rearranging:

\[
x^4 - 9x^2 - 4x^2 + 36 = 0
\]

This simplifies to:

\[
x^4 - 13x^2 + 36 = 0
\]

Let \(y = x^2\):

\[
y^2 - 13y + 36 = 0
\]

Factoring gives:

\[
(y-9)(y-4)=0 \Rightarrow y=9 \text{ or } y=4
\]

Thus, \(x^2 = 9 \Rightarrow x = 3, -3\) and \(x^2 = 4 \Rightarrow x = 2, -2\).

### 3) \(x^2(x+2) - 9x - 18 = 0\)

Expanding:

\[
x^3 + 2x^2 - 9x - 18 = 0
\]

This can be solved (possibly via synthetic division or numerical methods).

### 4) \(6(1-x) + x^2 - 2x + 1 = 0\)

Expanding:

\[
6 - 6x + x^2 - 2x + 1 = 0
\]

Combining like terms gives:

\[
x^2 - 8x + 7 = 0
\]

Factoring or using the quadratic formula:

\[
(x-7)(x-1)=0 \Rightarrow x=7 \text{ or } x=1
\]

Now, you can solve each cubic or quartic as necessary. If you need further detail on any specific equation, let me know!