What is expand y^{1/3}(y^{4/9}-7y^{2/9}) ?

The solution to expand y^{1/3}(y^{4/9}-7y^{2/9}) is y^{7/9}-7y^{5/9}

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expand y^{1/3}(y^{4/9}-7y^{2/9})

expand

y^{\frac{1}{3}} (y^{\frac{4}{9}} - 7 y^{\frac{2}{9}})

y^{\frac{7}{9}} - 7 y^{\frac{5}{9}}

Solution steps

y^{\frac{1}{3}} (y^{\frac{4}{9}} - 7 y^{\frac{2}{9}})

Apply the distributive law:

a (b - c) = ab - a c

y^{\frac{1}{3}} (y^{\frac{4}{9}} - 7 y^{\frac{2}{9}}) = y^{\frac{1}{3}} y^{\frac{4}{9}} - y^{\frac{1}{3}} \cdot 7 y^{\frac{2}{9}}

= y^{\frac{1}{3}} y^{\frac{4}{9}} - y^{\frac{1}{3}} \cdot 7 y^{\frac{2}{9}}

y^{\frac{1}{3}} y^{\frac{4}{9}} = y^{\frac{7}{9}}

y^{\frac{1}{3}} \cdot 7 y^{\frac{2}{9}} = 7 y^{\frac{5}{9}}

= y^{\frac{7}{9}} - 7 y^{\frac{5}{9}}

e^{x} cos (x) (e^{x} cos (x) + e^{x} sin (x))

\frac{x}{( x - 2 ) ( x + 2 )}

3 cos (t) sin (t) - (2 t - 5)^{2} sin (t) + 3 t

f (x) = \frac{1}{( x - 1 ) ^{5}}

- (x - 2) e^{2 x}

What is expand y^{1/3}(y^{4/9}-7y^{2/9}) ?
The solution to expand y^{1/3}(y^{4/9}-7y^{2/9}) is y^{7/9}-7y^{5/9}