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

The solution to expand (2x-1/(x^2))^4 is 16x^4-32x+(24)/(x^2)-8/(x^5)+1/(x^8)

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expand (2x-1/(x^2))^4

expand

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

16 x^{4} - 32 x + \frac{24}{x ^{2}} - \frac{8}{x ^{5}} + \frac{1}{x ^{8}}

Solution steps

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

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 2 x, b = - \frac{1}{x ^{2}}

= i = 0 \sum 4 (i 4) (2 x)^{(4 - i)} (- \frac{1}{x ^{2}})^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (2 x)^{4} (- \frac{1}{x ^{2}})^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (2 x)^{3} (- \frac{1}{x ^{2}})^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (2 x)^{2} (- \frac{1}{x ^{2}})^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (2 x)^{1} (- \frac{1}{x ^{2}})^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (2 x)^{0} (- \frac{1}{x ^{2}})^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (2 x)^{4} (- \frac{1}{x ^{2}})^{0} : 16 x^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (2 x)^{3} (- \frac{1}{x ^{2}})^{1} : - 32 x

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (2 x)^{2} (- \frac{1}{x ^{2}})^{2} : \frac{24}{x ^{2}}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (2 x)^{1} (- \frac{1}{x ^{2}})^{3} : - \frac{8}{x ^{5}}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (2 x)^{0} (- \frac{1}{x ^{2}})^{4} : \frac{1}{x ^{8}}

= 16 x^{4} - 32 x + \frac{24}{x ^{2}} - \frac{8}{x ^{5}} + \frac{1}{x ^{8}}

\frac{- x ^{2} + 5 x - 2}{x ( x + 1 ) ( x - 1 )}

\frac{x ( 10 x - x ^{2} )}{20}

4 ((k + 1) (k + 2) (k + 3))

(1 - x^{2}) y^{''} (x) - 10 x y^{'} (x) - 6 y

(x + 1) (x - x^{2})

What is expand (2x-1/(x^2))^4 ?
The solution to expand (2x-1/(x^2))^4 is 16x^4-32x+(24)/(x^2)-8/(x^5)+1/(x^8)