What is x^{-1/3}(x^{7/6}-x^{3/4}) ?

The solution to x^{-1/3}(x^{7/6}-x^{3/4}) is (x^{7/6}-x^{3/4})/(x^{1/3)}

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simplify x^{-1/3}(x^{7/6}-x^{3/4})

simplify

x^{- \frac{1}{3}} (x^{\frac{7}{6}} - x^{\frac{3}{4}})

\frac{x ^{\frac{7}{6}} - x ^{\frac{3}{4}}}{x ^{\frac{1}{3}}}

Solution steps

x^{- \frac{1}{3}} (x^{\frac{7}{6}} - x^{\frac{3}{4}})

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

x^{- \frac{1}{3}} = \frac{1}{x ^{\frac{1}{3}}}

= \frac{1}{x ^{\frac{1}{3}}} (x^{\frac{7}{6}} - x^{\frac{3}{4}})

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( x ^{\frac{7}{6}} - x ^{\frac{3}{4}} )}{x ^{\frac{1}{3}}}

Apply rule:

1 \cdot a = a

1 \cdot (x^{\frac{7}{6}} - x^{\frac{3}{4}}) = (x^{\frac{7}{6}} - x^{\frac{3}{4}})

= \frac{x ^{\frac{7}{6}} - x ^{\frac{3}{4}}}{x ^{\frac{1}{3}}}

\frac{1}{7} \cdot \frac{2}{9}

2 x^{2} + 9 x - 18

- 1 - 8 i - 4 - i

5 x^{2} - 9 x + 2 = 0

18 x y^{2} + 12 x y + 2 x

What is x^{-1/3}(x^{7/6}-x^{3/4}) ?
The solution to x^{-1/3}(x^{7/6}-x^{3/4}) is (x^{7/6}-x^{3/4})/(x^{1/3)}