Question

Solve the following equations,$\sqrt[3]{3x^2}-\sqrt[3]{3x}-6=0$

Answer 1

$\sqrt[3]{3x^2}-\sqrt[3]{3x}-6=0$

$\Rightarrow x^{2/3}-x^{1/3}=6⟶\left(1\right)$

taking cube on both sides

$\Rightarrow {(x^{2/3}-x^{1/3})}^3=6^3$

$\Rightarrow {\left(x^{2/3}\right)}^3-{\left(x^{1/3}\right)}^3-3x^{2/3}.x^{1/3}(x^{2/3}-x^{1/3})=216$

$\Rightarrow x^2-x-3x.6=216$ $(\therefore x^{2/3}-x^{1/3}=6)$

$\Rightarrow x^2-19x-216=0$

$\Rightarrow x=\frac{+19\pm \sqrt{{19}^2-4\times 1\times -216}}{2}$

$=\frac{19\pm \sqrt{361+864}}{2}$

$\Rightarrow x=\frac{19\pm \sqrt{1225}}{4}$

$=\frac{19\pm 15}{4}$

$\Rightarrow x=\frac{19+15}{4}=\frac{17}{2}$ or $x=\frac{19-15}{4}=2.$

Hence, solve.