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

The solution to expand 1/((3-x)^{-4)} is x^4-12x^3+54x^2-108x+81

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

expand

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

x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81

Solution steps

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

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

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

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply the fraction rule:

\frac{a}{1} = a

= (3 - x)^{4}

Apply binomial theorem:

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

a = 3, b = - x

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

Expand summation

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

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 3^{4} (- x)^{0} = 81

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 3^{3} (- x)^{1} : - 108 x

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 3^{2} (- x)^{2} : 54 x^{2}

Simplify

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

Simplify

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

= 81 - 108 x + 54 x^{2} - 12 x^{3} + x^{4}

Rewrite in standard form

= x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81

(- 1 - i 3)^{10}

(x + i y)^{8}

(x + i y)^{7}

\frac{1}{( 1 - e ^{- π s} ) ( s ^{2} + 1 )}

(1 + x y) e^{x y}

What is expand 1/((3-x)^{-4)} ?
The solution to expand 1/((3-x)^{-4)} is x^4-12x^3+54x^2-108x+81