What is expand (1/(x^2)+y^3)^4 ?

The solution to expand (1/(x^2)+y^3)^4 is 1/(x^8)+(4y^3)/(x^6)+(6y^6)/(x^4)+(4y^9)/(x^2)+y^{12}

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expand (1/(x^2)+y^3)^4

expand

(\frac{1}{x ^{2}} + y^{3})^{4}

\frac{1}{x ^{8}} + \frac{4 y ^{3}}{x ^{6}} + \frac{6 y ^{6}}{x ^{4}} + \frac{4 y ^{9}}{x ^{2}} + y^{12}

Solution steps

(\frac{1}{x ^{2}} + y^{3})^{4}

Apply binomial theorem:

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

a = \frac{1}{x ^{2}}, b = y^{3}

= i = 0 \sum 4 (i 4) (\frac{1}{x ^{2}})^{(4 - i)} (y^{3})^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (\frac{1}{x ^{2}})^{4} (y^{3})^{0} : \frac{1}{x ^{8}}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (\frac{1}{x ^{2}})^{3} (y^{3})^{1} : \frac{4 y ^{3}}{x ^{6}}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (\frac{1}{x ^{2}})^{2} (y^{3})^{2} : \frac{6 y ^{6}}{x ^{4}}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (\frac{1}{x ^{2}})^{1} (y^{3})^{3} : \frac{4 y ^{9}}{x ^{2}}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (\frac{1}{x ^{2}})^{0} (y^{3})^{4} : y^{12}

= \frac{1}{x ^{8}} + \frac{4 y ^{3}}{x ^{6}} + \frac{6 y ^{6}}{x ^{4}} + \frac{4 y ^{9}}{x ^{2}} + y^{12}

(1 + 2 x^{3}) e^{x^{2}}

- 9 (5 + 3 h - 4 i)

(x - \frac{2}{x})^{15}

l im x \Rightarrow 1 (x^{3} + 3 x^{2} - x - 3) (x^{2} - 1)

(z - \frac{1}{2}) (z - \frac{1}{3})

What is expand (1/(x^2)+y^3)^4 ?
The solution to expand (1/(x^2)+y^3)^4 is 1/(x^8)+(4y^3)/(x^6)+(6y^6)/(x^4)+(4y^9)/(x^2)+y^{12}