Respuesta :
[tex]\bf x^{\frac{2}{3}}-x^{\frac{1}{3}}+4=6\implies x^{\frac{2}{3}}-x^{\frac{1}{3}}-2=0
\\\\\\
\left( x^{\frac{1}{3}} \right)^2-\left( x^{\frac{1}{3}} \right)-2=0\impliedby \textit{now, that's just a quadratic}[/tex]
[tex]\bf \left( x^{\frac{1}{3}}-2 \right)\left( x^{\frac{1}{3}}+1 \right)=0\implies \begin{cases} x^{\frac{1}{3}}-2=0\\ x^{\frac{1}{3}}=2\\ \left( x^{\frac{1}{3}} \right)^3=2^3\\\\ \boxed{x=8}\\ ----------\\ x^{\frac{1}{3}}+1=0\\ x^{\frac{1}{3}}=-1\\ \left( x^{\frac{1}{3}} \right)^3=(-1)^3\\\\ \boxed{x=-1} \end{cases}[/tex]
[tex]\bf \left( x^{\frac{1}{3}}-2 \right)\left( x^{\frac{1}{3}}+1 \right)=0\implies \begin{cases} x^{\frac{1}{3}}-2=0\\ x^{\frac{1}{3}}=2\\ \left( x^{\frac{1}{3}} \right)^3=2^3\\\\ \boxed{x=8}\\ ----------\\ x^{\frac{1}{3}}+1=0\\ x^{\frac{1}{3}}=-1\\ \left( x^{\frac{1}{3}} \right)^3=(-1)^3\\\\ \boxed{x=-1} \end{cases}[/tex]
