To simplify the rational expression \(\frac{27 - 125r^3}{75r^2 + 45r + 27}\), we will factor both the numerator and the denominator.
Step 1: Factor the numerator \(27 - 125r^3\)
The numerator \(27 - 125r^3\) can be recognized as a difference of cubes. We can rewrite it as:
\[ 27 - (5r)^3 \]
Using the difference of cubes formula, \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), where \(a = 3\) and \(b = 5r\), we have:
\[ 27 - 125r^3 = (3 - 5r)((3)^2 + (3)(5r) + (5r)^2) = (3 - 5r)(9 + 15r + 25r^2) \]
So, the factored form of the numerator is:
\[ 27 - 125r^3 = (3 - 5r)(9 + 15r + 25r^2) \]
Step 2: Factor the denominator \(75r^2 + 45r + 27\)
To factor \(75r^2 + 45r + 27\), we can look for two numbers that multiply to \(75 \cdot 27\) (which is \(2025\)) and add to \(45\). The factors of \(75\) and \(27\) can help here:
This quadratic can be factored by using the method of grouping or attempting to directly factor. Factoring gives:
\[ 75r^2 + 45r + 27 = (15r + 9)(5r + 3) \]
Step 3: Rewrite the expression in lowest terms
Now, rewriting the entire expression with the factored forms gives us:
\[ \frac{(3 - 5r)(9 + 15r + 25r^2)}{(15r + 9)(5r + 3)} \]
Next, we can recognize that \(15r + 9\) can be factored further as:
\[ 15r + 9 = 3(5r + 3) \]
So the denominator can be rewritten as:
\[ (15r + 9)(5r + 3) = 3(5r + 3)(5r + 3) \]
Now we have:
\[ \frac{(3 - 5r)(9 + 15r + 25r^2)}{3(5r + 3)(5r + 3)} \]
Step 4: Cancel any common factors
We see we can cancel a \(3\) in the numerator and denominator:
\[ \frac{(3 - 5r)(9 + 15r + 25r^2)}{(5r + 3)(5r + 3)} \]
This leaves us with:
\[ \frac{3 - 5r}{(5r + 3)^2}(9 + 15r + 25r^2) \]
Final result:
The simplified expression is:
\[ \frac{(3 - 5r)(9 + 15r + 25r^2)}{(5r + 3)^2} \]
This is the lowest terms of the given rational expression.