Where is the step that caused the extraneous solution? $$x^{1/3}+x^{1/6}-2=0\iff (x^{1/6}-1)(x^{1/6}+2)=0$$ $\iff x^{1/6}=1$ or $x^{1/6}=-2$ $\implies x=1^6=1$ or $x=(-2)^6=64$ By checking when $x=64$:$\sqrt[3]{64}+\sqrt[6]{64}-2=4+2-2=4\ne0$ Which step caused the **extraneous solution** $x=64$ to appear

An even root cannot have negative values so the solution $\sqrt[6]{x}=-2$ is not valid in $\mathbb{R}$