Respuesta :
The equation is given to be:
[tex]125x^3-27=0[/tex]Applying the exponent rules, we have:
[tex]\Rightarrow(5x)^3-3^3=0[/tex]Recall the Difference of Cubes formula given to be:
[tex]x^3-y^3=\mleft(x-y\mright)\mleft(x^2+xy+y^2\mright)[/tex]Therefore, the equation becomes:
[tex]\begin{gathered} (5x-3)(\lbrack5x\rbrack^2+\lbrack3\times5x\rbrack+3^2)=0 \\ (5x-3)(25x^2+15x+9)=0 \end{gathered}[/tex]Using the Zero Factor Principle, we have that:
[tex]\begin{gathered} (x-a)(x-b)=0 \\ \text{then} \\ x-a=0 \\ \text{and} \\ x-b=0 \end{gathered}[/tex]Therefore, we can have:
[tex]\begin{gathered} 5x-3=0 \\ 5x=3 \\ x=\frac{3}{5} \end{gathered}[/tex]and
[tex]25x^2+15x+9=0[/tex]Solving the equation using the Quadratic Formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]From the Quadratic Equation, we have:
[tex]\begin{gathered} a=25 \\ b=15 \\ c=9 \end{gathered}[/tex]Therefore, we have:
[tex]x=\frac{-15\pm\sqrt[]{15^2-\lbrack4\times25\times9\rbrack}}{2\times25}=\frac{-15\pm\sqrt[]{225-900}}{50}=\frac{-15\pm\sqrt[]{-675}}{50}[/tex]Recall that:
[tex]\sqrt[]{-1}=i[/tex]Therefore, we have that:
[tex]x=\frac{-15\pm\sqrt[]{-1\times675}}{50}=\frac{-15\pm i\sqrt[]{225\times3}}{50}=\frac{-15\pm15i\sqrt[]{3}}{50}[/tex]Hence, the values of x can be:
[tex]x=\frac{-15_{}+15i\sqrt[]{3}}{50}=\frac{5(-3+3i\sqrt[]{3})}{50}=\frac{-3+3i\sqrt[]{3}}{10}=-\frac{3}{10}+i\frac{3\sqrt[]{3}}{10}[/tex]or
[tex]x=\frac{-15_{}-15i\sqrt[]{3}}{50}=\frac{5(-3-3i\sqrt[]{3})}{50}=\frac{-3-3i\sqrt[]{3}}{10}=-\frac{3}{10}-i\frac{3\sqrt[]{3}}{10}[/tex]ANSWER:
[tex]\begin{gathered} x=\frac{3}{5} \\ x=-\frac{3}{10}+i\frac{3\sqrt[]{3}}{10} \\ x=-\frac{3}{10}-i\frac{3\sqrt[]{3}}{10} \end{gathered}[/tex]