What is expand (1-x^4)^7 ?

The solution to expand (1-x^4)^7 is 1-7x^4+21x^8-35x^{12}+35x^{16}-21x^{20}+7x^{24}-x^{28}

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expand (1-x^4)^7

expand

(1 - x^{4})^{7}

1 - 7 x^{4} + 21 x^{8} - 35 x^{12} + 35 x^{16} - 21 x^{20} + 7 x^{24} - x^{28}

Solution steps

(1 - x^{4})^{7}

Apply binomial theorem:

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

a = 1, b = - x^{4}

= i = 0 \sum 7 (i 7) \cdot 1^{(7 - i)} (- x^{4})^{i}

Expand summation

= \frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 1^{7} (- x^{4})^{0} + \frac{7 !}{1 ! ( 7 - 1 ) !} \cdot 1^{6} (- x^{4})^{1} + \frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 1^{5} (- x^{4})^{2} + \frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 1^{4} (- x^{4})^{3} + \frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 1^{3} (- x^{4})^{4} + \frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 1^{2} (- x^{4})^{5} + \frac{7 !}{6 ! ( 7 - 6 ) !} \cdot 1^{1} (- x^{4})^{6} + \frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 1^{0} (- x^{4})^{7}

Simplify

\frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 1^{7} (- x^{4})^{0} : 1

Simplify

\frac{7 !}{1 ! ( 7 - 1 ) !} \cdot 1^{6} (- x^{4})^{1} : - 7 x^{4}

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 1^{5} (- x^{4})^{2} : 21 x^{8}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 1^{4} (- x^{4})^{3} : - 35 x^{12}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 1^{3} (- x^{4})^{4} : 35 x^{16}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 1^{2} (- x^{4})^{5} : - 21 x^{20}

Simplify

\frac{7 !}{6 ! ( 7 - 6 ) !} \cdot 1^{1} (- x^{4})^{6} : 7 x^{24}

Simplify

\frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 1^{0} (- x^{4})^{7} : - x^{28}

= 1 - 7 x^{4} + 21 x^{8} - 35 x^{12} + 35 x^{16} - 21 x^{20} + 7 x^{24} - x^{28}

8 x^{6} + 125

\frac{l ^{- 1} \cdot s}{( ( s ^{2} + 1 ) ( s ^{2} + 4 ) ( s ^{2} + 9 ) )}

- 1 x y + 5 x y^{2} z

343 x^{3} - 1331 y^{3}

(a + b)^{3} - 3 a \cdot b (a + b)

What is expand (1-x^4)^7 ?
The solution to expand (1-x^4)^7 is 1-7x^4+21x^8-35x^{12}+35x^{16}-21x^{20}+7x^{24}-x^{28}