Domains of fractional-power substitutions Interpret real fractional powers through roots. x1/3= [3]x permits negative x; x1/6= [6]x requires x 0 and is nonnegative. The substituted expression has its own range. For u=x2/3=([3]x)2, u 0, and each u>0 allows both signs of [3]x. State the original domain and the substitution range; solve the quadratic; reject unattainable substituted roots; reverse the actual root/power relation carefully; check each real answer. Example For x1/3-3x1/6+2=0, set u=x1/6 0. (u-1)(u-2)=0, so x=16=1 or x=26=64. Check: x2/3=4 means [3]x=+-2, hence x=+-8. It does not mean x=8 alone. A principal square or sixth root cannot be negative.

Domains of fractional-power substitutions

Interpret real fractional powers through roots. \(x^{\frac{1}{3}}=\sqrt[3]{x}\) permits negative \(x\); \(x^{\frac{1}{6}}=\sqrt[6]{x}\) requires \(x\ge0\) and is nonnegative.

The substituted expression has its own range. For \(u=x^{\frac{2}{3}}=(\sqrt[3]{x})^2\), \(u\ge0\), and each \(u>0\) allows both signs of \(\sqrt[3]{x}\).

State the original domain and the substitution range; solve the quadratic; reject unattainable substituted roots; reverse the actual root/power relation carefully; check each real answer.

For \(x^{\frac{1}{3}}-3x^{\frac{1}{6}}+2=0\), set \(u=x^{\frac{1}{6}}\ge0\). \((u-1)(u-2)=0\), so \(x=1^6=1\) or \(x=2^6=64\).

Check: \(x^{\frac{2}{3}}=4\) means \(\sqrt[3]{x}=\pm2\), hence \(x=\pm8\). It does not mean \(x=8\) alone. A principal square or sixth root cannot be negative.