Question

Solve for the roots in the equation below. In your final answer, include each of the necessary steps and calculations. Hint: Use your knowledge of factoring polynomials and identities for imaginary numbers.

x^3- 27i = 0

Answer 1

`\bf 27\implies 3^3\\\\ i^3\implies i\cdot i\cdot i\implies i^2\cdot i\implies -1\cdot i\implies -i\\\\ -----------------------------\\\\ x^3-27i=0\implies x^3+3^3(-i)=0\implies x^3+(3^3i^3)=0 \\\\\\ x^3+(3i)^3=0\\\\ -----------------------------\\\\ \textit{difference of cubes} \\ \quad \\ a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad (a+b)(a^2-ab+b^2)= a^3+b^3\\\\ -----------------------------\\\\`


`\bf (x+3i)[x^2-3ix+(3i)^2]=0\implies \begin{cases} x+3i=0\implies \boxed{x=-3i}\\\\ x^2-3ix+(3i)^2=0 \end{cases} \\\\\\ \textit{now, for the second one, we'd need the quadratic formula} \\\\\\ x^2-3ix+(3i)^2=0\implies x^2-3ix+(3^2i^2)=0`


`\bf x^2-3ix+(-9)=0\implies x^2-3ix-9=0 \\\\\\ \textit{quadratic formula}\\\\ x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}\implies x=\cfrac{3i\pm√((-3i)^2-4(1)(-9))}{2(1)} \\\\\\ x=\cfrac{3i\pm√((-3)^2i^2+36)}{2}\implies x=\cfrac{3i\pm√(-9+36)}{2} \\\\\\ x=\cfrac{3i\pm√(27)}{2}\implies \boxed{x=\cfrac{3i\pm 3√(3)}{2}}`