Question

The binomial (2x3 + 3y2)7 is shown in expanded form, but elements of several terms are missing. To complete the expansion, drag the missing parts to the places they belong.

Answer 1

${(2x^{3} + 3x^{2} )}^{7}$
EXPANDED: VERY LONG

$128x^{21} +1344x^{18} y^{2} +6048x^{15} y^{4} +15120x^{12} y^{6} +22680x^{9} y^{8} +20412x^{6} y^{10} +10206x^{3} y^{12} +2187y^{14}$

I used the Binomial Theorem, Which States
$(a + b)(a + b)= {a}^{2} + 2ab + {b}^{2}$
$({a}^{2} + 2ab + {b}^{2} )(a + b)={a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3}$
$({a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3}) \times (a + b) = i \: believe \: you \: get \: the \: pattern \: now$

Answer 2

$\left(2 x^{3} + 3 y^{2}\right)^{7} = 128 x^{21} + 1344 x^{18} y^{2} + 6048 x^{15} y^{4} + 15120 x^{12} y^{6} + 22680 x^{9} y^{8} + 20412 x^{6} y^{10} + 10206 x^{3} y^{12} + 2187 y^{14}.$

The binomial expansion is given by the 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 $n! = 1 \cdot 2 \cdot \ldots \cdot n$.

Here, $\left(2 x^{3} + 3 y^{2}\right)^{7} = \sum_{k=0}^{7} {\binom{7}{k}} \left(2 x^{3}\right)^{7 - k} \left(3 y^{2}\right)^{k}$

calculate the product for every value of $k$ from $0$ to $7$.

For $k=0$, ${\binom{7}{0}} \left(2 x^{3}\right)^{7 - 0} \left(3 y^{2}\right)^{0} = \frac{7!}{\left(7 - 0\right)! 0!} \left(2 x^{3}\right)^{7 - 0} \left(3 y^{2}\right)^{0} = 128 x^{21}$

For $k=1$, ${\binom{7}{1}} \left(2 x^{3}\right)^{7 - 1} \left(3 y^{2}\right)^{1} = \frac{7!}{\left(7 - 1\right)! 1!} \left(2 x^{3}\right)^{7 - 1} \left(3 y^{2}\right)^{1} = 1344 x^{18} y^{2}$

For $k=2$, ${\binom{7}{2}} \left(2 x^{3}\right)^{7 - 2} \left(3 y^{2}\right)^{2} = \frac{7!}{\left(7 - 2\right)! 2!} \left(2 x^{3}\right)^{7 - 2} \left(3 y^{2}\right)^{2} = 6048 x^{15} y^{4}$

For $k=3$, ${\binom{7}{3}} \left(2 x^{3}\right)^{7 - 3} \left(3 y^{2}\right)^{3} = \frac{7!}{\left(7 - 3\right)! 3!} \left(2 x^{3}\right)^{7 - 3} \left(3 y^{2}\right)^{3} = 15120 x^{12} y^{6}$

For $k=4$, ${\binom{7}{4}} \left(2 x^{3}\right)^{7 - 4} \left(3 y^{2}\right)^{4} = \frac{7!}{\left(7 - 4\right)! 4!} \left(2 x^{3}\right)^{7 - 4} \left(3 y^{2}\right)^{4} = 22680 x^{9} y^{8}$

For $k=5$, ${\binom{7}{5}} \left(2 x^{3}\right)^{7 - 5} \left(3 y^{2}\right)^{5} = \frac{7!}{\left(7 - 5\right)! 5!} \left(2 x^{3}\right)^{7 - 5} \left(3 y^{2}\right)^{5} = 20412 x^{6} y^{10}$

For $k=6$, ${\binom{7}{6}} \left(2 x^{3}\right)^{7 - 6} \left(3 y^{2}\right)^{6} = \frac{7!}{\left(7 - 6\right)! 6!} \left(2 x^{3}\right)^{7 - 6} \left(3 y^{2}\right)^{6} = 10206 x^{3} y^{12}$

For $k=7$, ${\binom{7}{7}} \left(2 x^{3}\right)^{7 - 7} \left(3 y^{2}\right)^{7} = \frac{7!}{\left(7 - 7\right)! 7!} \left(2 x^{3}\right)^{7 - 7} \left(3 y^{2}\right)^{7} = 2187 y^{14}$

So, $\left(2 x^{3} + 3 y^{2}\right)^{7} = 128 x^{21} + 1344 x^{18} y^{2} + 6048 x^{15} y^{4} + 15120 x^{12} y^{6} + 22680 x^{9} y^{8} + 20412 x^{6} y^{10} + 10206 x^{3} y^{12} + 2187 y^{14}.$

Learn more about binomial expansion here:

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