Question

solve for the roots in the equation below. in your final answer include all of the necessary steps and calculations. x^3-27 =0

Answer 1

`\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x^3-27=0\implies x^3-3^3=0\implies (x-3)(x^2+3x+9)=0 \\\\\\ x-3=0\implies \boxed{x=3}`

now, we could also solve by the other factor, the trinomial x² + 3x + 9 = 0, however, you can check that in the quadratic formula is you wish, we'd end up with "imaginary roots", or namely complex roots, which is another way of saying, no solution, or no real roots from that factor.