Question

the first four terms of the expansion of the binomial \\((2x^3 + 3y^2)^7\\) are shown.
\\(128x^{21} + 1,344x^{18}y^2 + 6,048x^{15}y^4 + 15,120x^{12}y^6 + \dots\\)
complete the last four terms by dragging the missing values to the correct locations in the expression. not all values will be used.
\\(\dots + \boxed{}x^9y^8 + 20,412\boxed{} + \boxed{}y^{12} + \boxed{}\\)
\\(22,680\quad x^6\quad x^7y^7\quad 10,206\quad 1,344\\)
\\(2,187y^{14}\quad 3y^7\quad x^3\quad x^6y^{10}\\)

Explanation:

Step1: Recall Binomial Theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\), \(a = 2x^{3}\), \(b=3y^{2}\), and \(n = 7\).

Step2: Find the general term

The general term \(T_{k+1}=\binom{7}{k}(2x^{3})^{7 - k}(3y^{2})^{k}=\binom{7}{k}2^{7 - k}3^{k}x^{21 - 3k}y^{2k}\)

Step3: Find the 5th term (\(k = 4\))

For \(k = 4\):
\(\binom{7}{4}=\frac{7!}{4!3!}=35\)
\(2^{7 - 4}=2^{3}=8\)
\(3^{4}=81\)
\(T_{5}=35\times8\times81x^{21-12}y^{8}=35\times648x^{9}y^{8}=22680x^{9}y^{8}\)

Step4: Find the 6th term (\(k = 5\))

For \(k = 5\):
\(\binom{7}{5}=\frac{7!}{5!2!}=21\)
\(2^{7 - 5}=2^{2}=4\)
\(3^{5}=243\)
\(T_{6}=21\times4\times243x^{21 - 15}y^{10}=21\times972x^{6}y^{10}=20412x^{6}y^{10}\) (matches the given coefficient 20412, so the missing part is \(x^{6}y^{10}\))

Step5: Find the 7th term (\(k = 6\))

For \(k = 6\):
\(\binom{7}{6}=\frac{7!}{6!1!}=7\)
\(2^{7 - 6}=2^{1}=2\)
\(3^{6}=729\)
\(T_{7}=7\times2\times729x^{21-18}y^{12}=7\times1458x^{3}y^{12}=10206x^{3}y^{12}\)

Step6: Find the 8th term (\(k = 7\))

For \(k = 7\):
\(\binom{7}{7}=\frac{7!}{7!0!}=1\)
\(2^{7 - 7}=2^{0}=1\)
\(3^{7}=2187\)
\(T_{8}=1\times1\times2187x^{21-21}y^{14}=2187y^{14}\)

Answer:

... + \(\boldsymbol{22680}\) \(x^{9}y^{8}\) + 20,412 \(\boldsymbol{x^{6}y^{10}}\) + \(\boldsymbol{10206}\) \(x^{3}y^{12}\) + \(\boldsymbol{2187y^{14}}\)