Question

the first four terms of the expansion of the binomial ((2x^3 + 3y^2)^7) are shown: (128x^{21}+1,344x^{18}y^3 + 6,048x^{15}y^6+15,120x^{12}y^9+cdots) complete the last four terms by dragging the missing values to the correct locations in the expression. not all values will be used. (cdots + quadx^9y^{12}+20,412quad+quadquady^{18}+quad) the options are: (x^7y^7), (x^3), (1,344), (22,680), (3y^2), (2,187y^{14}), (10,206), (x^6y^{10}), (x^0)

Explanation:

Step1: Recall binomial theorem formula

The binomial expansion formula is:
$$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$$
Here, $a=2x^3$, $b=3y^2$, $n=7$.

Step2: Calculate 5th term (k=4)

Find coefficient, $x$, $y$ exponents:
$\binom{7}{4}(2x^3)^{7-4}(3y^2)^{4}=35\times(2^3x^9)\times(3^4y^8)=35\times8\times81x^9y^8=22680x^9y^8$

Step3: Calculate 6th term (k=5)

Find coefficient, $x$, $y$ exponents:
$\binom{7}{5}(2x^3)^{7-5}(3y^2)^{5}=21\times(2^2x^6)\times(3^5y^{10})=21\times4\times243x^6y^{10}=20412x^6y^{10}$

Step4: Calculate 7th term (k=6)

Find coefficient, $x$, $y$ exponents:
$\binom{7}{6}(2x^3)^{7-6}(3y^2)^{6}=7\times(2x^3)\times(3^6y^{12})=7\times2\times729x^3y^{12}=10206x^3y^{12}$

Step5: Calculate 8th term (k=7)

Find coefficient, $x$, $y$ exponents:
$\binom{7}{7}(2x^3)^{7-7}(3y^2)^{7}=1\times1\times(3^7y^{14})=2187y^{14}$

Answer:

$\dots + \boldsymbol{22,680}x^9y^8 + 20,412\boldsymbol{x^6y^{10}} + \boldsymbol{10,206x^3}y^{12} + \boldsymbol{2,187y^{14}}$