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

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

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expand (1-x^3)^4

expand

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

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

Solution steps

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

Apply binomial theorem:

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

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

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

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

\frac{( 8 x + 64 y )}{( - 4 )}

5 x y - 15 x^{3} y^{3}

x^{3} - 11 x^{2} + 36 x - 36

(b) {(\frac{1}{2})^{∣} k ∣}

(\frac{x - y}{3})^{3}

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