What is expand (x^2-1)^8 ?

The solution to expand (x^2-1)^8 is x^{16}-8x^{14}+28x^{12}-56x^{10}+70x^8-56x^6+28x^4-8x^2+1

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expand (x^2-1)^8

expand

(x^{2} - 1)^{8}

x^{16} - 8 x^{14} + 28 x^{12} - 56 x^{10} + 70 x^{8} - 56 x^{6} + 28 x^{4} - 8 x^{2} + 1

Solution steps

(x^{2} - 1)^{8}

Apply binomial theorem:

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

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

= i = 0 \sum 8 (i 8) (x^{2})^{(8 - i)} (- 1)^{i}

Expand summation

= \frac{8 !}{0 ! ( 8 - 0 ) !} (x^{2})^{8} (- 1)^{0} + \frac{8 !}{1 ! ( 8 - 1 ) !} (x^{2})^{7} (- 1)^{1} + \frac{8 !}{2 ! ( 8 - 2 ) !} (x^{2})^{6} (- 1)^{2} + \frac{8 !}{3 ! ( 8 - 3 ) !} (x^{2})^{5} (- 1)^{3} + \frac{8 !}{4 ! ( 8 - 4 ) !} (x^{2})^{4} (- 1)^{4} + \frac{8 !}{5 ! ( 8 - 5 ) !} (x^{2})^{3} (- 1)^{5} + \frac{8 !}{6 ! ( 8 - 6 ) !} (x^{2})^{2} (- 1)^{6} + \frac{8 !}{7 ! ( 8 - 7 ) !} (x^{2})^{1} (- 1)^{7} + \frac{8 !}{8 ! ( 8 - 8 ) !} (x^{2})^{0} (- 1)^{8}

Simplify

\frac{8 !}{0 ! ( 8 - 0 ) !} (x^{2})^{8} (- 1)^{0} : x^{16}

Simplify

\frac{8 !}{1 ! ( 8 - 1 ) !} (x^{2})^{7} (- 1)^{1} : - 8 x^{14}

Simplify

\frac{8 !}{2 ! ( 8 - 2 ) !} (x^{2})^{6} (- 1)^{2} : 28 x^{12}

Simplify

\frac{8 !}{3 ! ( 8 - 3 ) !} (x^{2})^{5} (- 1)^{3} : - 56 x^{10}

Simplify

\frac{8 !}{4 ! ( 8 - 4 ) !} (x^{2})^{4} (- 1)^{4} : 70 x^{8}

Simplify

\frac{8 !}{5 ! ( 8 - 5 ) !} (x^{2})^{3} (- 1)^{5} : - 56 x^{6}

Simplify

\frac{8 !}{6 ! ( 8 - 6 ) !} (x^{2})^{2} (- 1)^{6} : 28 x^{4}

Simplify

\frac{8 !}{7 ! ( 8 - 7 ) !} (x^{2})^{1} (- 1)^{7} : - 8 x^{2}

Simplify

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

= x^{16} - 8 x^{14} + 28 x^{12} - 56 x^{10} + 70 x^{8} - 56 x^{6} + 28 x^{4} - 8 x^{2} + 1

15 x^{4} x^{10} x^{125}

6 (x - 0.5) (x + 3)

t + t^{- 1}

x^{4} - 3 x^{2} + 9

[a^{3}]^{\frac{3}{3}}

What is expand (x^2-1)^8 ?
The solution to expand (x^2-1)^8 is x^{16}-8x^{14}+28x^{12}-56x^{10}+70x^8-56x^6+28x^4-8x^2+1