Binomial Expansion Calculator

Expand $$$\left(2 x + 5\right)^{3}$$$.

The expansion is given by the following formula: $$$\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}$$$, where $$${\binom{n}{k}} = \frac{n!}{\left(n - k\right)! k!}$$$ and $$$n! = 1 \cdot 2 \cdot \ldots \cdot n$$$.

We have that $$$a = 2 x$$$, $$$b = 5$$$, and $$$n = 3$$$.

Therefore, $$$\left(2 x + 5\right)^{3} = \sum_{k=0}^{3} {\binom{3}{k}} \left(2 x\right)^{3 - k} 5^{k}$$$.

Now, calculate the product for every value of $$$k$$$ from $$$0$$$ to $$$3$$$.

$$$k = 0$$$: $$${\binom{3}{0}} \left(2 x\right)^{3 - 0} \cdot 5^{0} = \frac{3!}{\left(3 - 0\right)! 0!} \left(2 x\right)^{3 - 0} \cdot 5^{0} = 8 x^{3}$$$

$$$k = 1$$$: $$${\binom{3}{1}} \left(2 x\right)^{3 - 1} \cdot 5^{1} = \frac{3!}{\left(3 - 1\right)! 1!} \left(2 x\right)^{3 - 1} \cdot 5^{1} = 60 x^{2}$$$

$$$k = 2$$$: $$${\binom{3}{2}} \left(2 x\right)^{3 - 2} \cdot 5^{2} = \frac{3!}{\left(3 - 2\right)! 2!} \left(2 x\right)^{3 - 2} \cdot 5^{2} = 150 x$$$

$$$k = 3$$$: $$${\binom{3}{3}} \left(2 x\right)^{3 - 3} \cdot 5^{3} = \frac{3!}{\left(3 - 3\right)! 3!} \left(2 x\right)^{3 - 3} \cdot 5^{3} = 125$$$

Thus, $$$\left(2 x + 5\right)^{3} = 8 x^{3} + 60 x^{2} + 150 x + 125$$$.

$$$\left(2 x + 5\right)^{3} = 8 x^{3} + 60 x^{2} + 150 x + 125$$$A