What is expand (4(x^2-3))/(33(x^2-1)^{4/3)} ?

The solution to expand (4(x^2-3))/(33(x^2-1)^{4/3)} is (4x^2)/(33(x^2-1)^{4/3)}-4/(11(x^2-1)^{4/3)}

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expand (4(x^2-3))/(33(x^2-1)^{4/3)}

expand

\frac{4 ( x ^{2} - 3 )}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}}

\frac{4 x ^{2}}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}} - \frac{4}{11 ( x ^{2} - 1 ) ^{\frac{4}{3}}}

Solution steps

\frac{4 ( x ^{2} - 3 )}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}}

Expand

4 (x^{2} - 3) : 4 x^{2} - 12

= \frac{4 x ^{2} - 12}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}}

Apply the fraction rule:

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

\frac{4 x ^{2} - 12}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}} = \frac{4 x ^{2}}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}} - \frac{12}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}}

= \frac{4 x ^{2}}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}} - \frac{12}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}}

Cancel

\frac{12}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}} : \frac{4}{11 ( x ^{2} - 1 ) ^{\frac{4}{3}}}

= \frac{4 x ^{2}}{33 ( x ^{2} - 1 ) ^{\frac{4}{3}}} - \frac{4}{11 ( x ^{2} - 1 ) ^{\frac{4}{3}}}

n gu l a r (4, - \frac{π}{12})

\frac{7 e ^{2}}{( 7 + e ^{2} ) ^{2}}

(2 \cdot α) \cdot (A \cdot \frac{C}{2} + B \frac{C}{2})

\frac{( 9 x - 7 )}{( ( x + 3 ) ( x ^{(2)} - 1 ) )}

\frac{8 ( 4 - 13 )}{3}

What is expand (4(x^2-3))/(33(x^2-1)^{4/3)} ?
The solution to expand (4(x^2-3))/(33(x^2-1)^{4/3)} is (4x^2)/(33(x^2-1)^{4/3)}-4/(11(x^2-1)^{4/3)}