What is expand (4n^3+1)^4 ?

The solution to expand (4n^3+1)^4 is 256n^{12}+256n^9+96n^6+16n^3+1

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expand (4n^3+1)^4

expand

(4 n^{3} + 1)^{4}

256 n^{12} + 256 n^{9} + 96 n^{6} + 16 n^{3} + 1

Solution steps

(4 n^{3} + 1)^{4}

Apply binomial theorem:

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

a = 4 n^{3}, b = 1

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

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (4 n^{3})^{4} \cdot 1^{0} : 256 n^{12}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (4 n^{3})^{3} \cdot 1^{1} : 256 n^{9}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (4 n^{3})^{2} \cdot 1^{2} : 96 n^{6}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (4 n^{3})^{1} \cdot 1^{3} : 16 n^{3}

Simplify

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

= 256 n^{12} + 256 n^{9} + 96 n^{6} + 16 n^{3} + 1

120 x

\frac{( x ^{5} + 2 x ^{3} + 3 x )}{( x ^{2} + x )}

\frac{3 ^{- 1} + 4 ^{- 1}}{5 ^{- 1}}

2 (sin (x))^{2} + cos (2 x)

(3 - 2 i)^{10}

What is expand (4n^3+1)^4 ?
The solution to expand (4n^3+1)^4 is 256n^{12}+256n^9+96n^6+16n^3+1