What is expand (x^3-1)^4 ?

The solution to expand (x^3-1)^4 is x^{12}-4x^9+6x^6-4x^3+1

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expand (x^3-1)^4

expand

(x^{3} - 1)^{4}

x^{12} - 4 x^{9} + 6 x^{6} - 4 x^{3} + 1

Solution steps

(x^{3} - 1)^{4}

Apply binomial theorem:

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

a = x^{3}, b = - 1

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

= x^{12} - 4 x^{9} + 6 x^{6} - 4 x^{3} + 1

p^{4} + 4 p^{2} + 16

(2 x^{2} y^{- 3})^{- 2}

\frac{1}{25} + \frac{( 25 x ^{4} )}{36} - \frac{( x ^{2} )}{3}

4 y^{4} - 5 y^{2} + 1

6 x - 2 y + 7 - (2 x + 3)

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