use to solve \( 27 x^{3}-125=0 \) by factoring?

To solve \( 27 x^{3} - 125 = 0 \) by factoring, you can recognize that this is a difference of cubes. It can be expressed as \( (3x)^3 - 5^3 = 0 \). This difference of cubes can be factored using the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). In this case, let \( a = 3x \) and \( b = 5 \). Thus, it factors to: \[ (3x - 5)((3x)^2 + (3x)(5) + 5^2) = 0. \] Calculating the second part gives \( (3x)^2 + 15x + 25 \), leading to: \[ (3x - 5)(9x^2 + 15x + 25) = 0. \] Next, you can solve for \( x \) by setting the first factor to zero: \[ 3x - 5 = 0 \quad \Rightarrow \quad x = \frac{5}{3}. \] The second factor, \( 9x^2 + 15x + 25 \), does not have real roots (as its discriminant is negative). So, the only real solution is \( x = \frac{5}{3} \). Now you have your answer and a handy method for similar problems!