What is expand (2x^3+1)^5 ?

The solution to expand (2x^3+1)^5 is 32x^{15}+80x^{12}+80x^9+40x^6+10x^3+1

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expand (2x^3+1)^5

expand

(2 x^{3} + 1)^{5}

32 x^{15} + 80 x^{12} + 80 x^{9} + 40 x^{6} + 10 x^{3} + 1

Solution steps

(2 x^{3} + 1)^{5}

Apply binomial theorem:

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

a = 2 x^{3}, b = 1

= i = 0 \sum 5 (i 5) (2 x^{3})^{(5 - i)} \cdot 1^{i}

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} (2 x^{3})^{5} \cdot 1^{0} : 32 x^{15}

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (2 x^{3})^{4} \cdot 1^{1} : 80 x^{12}

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (2 x^{3})^{3} \cdot 1^{2} : 80 x^{9}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (2 x^{3})^{2} \cdot 1^{3} : 40 x^{6}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (2 x^{3})^{1} \cdot 1^{4} : 10 x^{3}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} (2 x^{3})^{0} \cdot 1^{5} : 1

= 32 x^{15} + 80 x^{12} + 80 x^{9} + 40 x^{6} + 10 x^{3} + 1

\frac{14}{35}

5 x + 5

\frac{x + 1}{x - 1} - \frac{x - 1}{x + 1}

lo g_{9} (5)

- 3 x - 6 + (- 1)

What is expand (2x^3+1)^5 ?
The solution to expand (2x^3+1)^5 is 32x^{15}+80x^{12}+80x^9+40x^6+10x^3+1