What is expand (2x^{5/3}(4t^2-x))/(t^{1/3)} ?

The solution to expand (2x^{5/3}(4t^2-x))/(t^{1/3)} is 8x^{5/3}t^{5/3}-(2x^{8/3})/(t^{1/3)}

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expand (2x^{5/3}(4t^2-x))/(t^{1/3)}

expand

\frac{2 x ^{\frac{5}{3}} ( 4 t ^{2} - x )}{t ^{\frac{1}{3}}}

8 x^{\frac{5}{3}} t^{\frac{5}{3}} - \frac{2 x ^{\frac{8}{3}}}{t ^{\frac{1}{3}}}

Solution steps

\frac{2 x ^{\frac{5}{3}} ( 4 t ^{2} - x )}{t ^{\frac{1}{3}}}

Expand

2 x^{\frac{5}{3}} (4 t^{2} - x) : 8 t^{2} x^{\frac{5}{3}} - 2 x^{\frac{8}{3}}

= \frac{8 t ^{2} x ^{\frac{5}{3}} - 2 x ^{\frac{8}{3}}}{t ^{\frac{1}{3}}}

Apply the fraction rule:

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

\frac{8 t ^{2} x ^{\frac{5}{3}} - 2 x ^{\frac{8}{3}}}{t ^{\frac{1}{3}}} = \frac{8 t ^{2} x ^{\frac{5}{3}}}{t ^{\frac{1}{3}}} - \frac{2 x ^{\frac{8}{3}}}{t ^{\frac{1}{3}}}

= \frac{8 t ^{2} x ^{\frac{5}{3}}}{t ^{\frac{1}{3}}} - \frac{2 x ^{\frac{8}{3}}}{t ^{\frac{1}{3}}}

Cancel

\frac{8 t ^{2} x ^{\frac{5}{3}}}{t ^{\frac{1}{3}}} : 8 x^{\frac{5}{3}} t^{\frac{5}{3}}

= 8 x^{\frac{5}{3}} t^{\frac{5}{3}} - \frac{2 x ^{\frac{8}{3}}}{t ^{\frac{1}{3}}}

\frac{2 ( x + 1 )}{( 3 x + 7 ) ^{\frac{1}{3}}}

\frac{6 x ^{2}}{( 1 - 9 x ) ^{2}}

(3 + i)^{8}

(\frac{x}{6} - \frac{y}{8}) (\frac{x}{6} + \frac{y}{8})

cos (3 (t - \frac{π}{2}))

What is expand (2x^{5/3}(4t^2-x))/(t^{1/3)} ?
The solution to expand (2x^{5/3}(4t^2-x))/(t^{1/3)} is 8x^{5/3}t^{5/3}-(2x^{8/3})/(t^{1/3)}