What is expand (x^{1/3}-1/4 x^{-1/3})^2 ?

The solution to expand (x^{1/3}-1/4 x^{-1/3})^2 is (16x^{4/3}-8x^{2/3}+1)/(16x^{2/3)}

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expand (x^{1/3}-1/4 x^{-1/3})^2

expand

(x^{\frac{1}{3}} - \frac{1}{4} x^{- \frac{1}{3}})^{2}

\frac{16 x ^{\frac{4}{3}} - 8 x ^{\frac{2}{3}} + 1}{16 x ^{\frac{2}{3}}}

Solution steps

(x^{\frac{1}{3}} - \frac{1}{4} x^{- \frac{1}{3}})^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

(x^{\frac{1}{3}} - \frac{1}{4} x^{- \frac{1}{3}})^{2} = (x^{\frac{1}{3}})^{2} - 2 x^{\frac{1}{3}} \frac{1}{4} x^{- \frac{1}{3}} + (\frac{1}{4} x^{- \frac{1}{3}})^{2}

= (x^{\frac{1}{3}})^{2} - 2 x^{\frac{1}{3}} \frac{1}{4} x^{- \frac{1}{3}} + (\frac{1}{4} x^{- \frac{1}{3}})^{2}

Simplify

(x^{\frac{1}{3}})^{2} - 2 x^{\frac{1}{3}} \frac{1}{4} x^{- \frac{1}{3}} + (\frac{1}{4} x^{- \frac{1}{3}})^{2} : \frac{16 x ^{\frac{4}{3}} - 8 x ^{\frac{2}{3}} + 1}{16 x ^{\frac{2}{3}}}

= \frac{16 x ^{\frac{4}{3}} - 8 x ^{\frac{2}{3}} + 1}{16 x ^{\frac{2}{3}}}

tan (h) (x - y)

(x + ln (x))^{2}

lo g_{9 (\frac{3 ^{3}}{7})^{4}} (x)

(x + 2) d x

e^{x y} (y) (x^{2} + y^{2})

What is expand (x^{1/3}-1/4 x^{-1/3})^2 ?
The solution to expand (x^{1/3}-1/4 x^{-1/3})^2 is (16x^{4/3}-8x^{2/3}+1)/(16x^{2/3)}