Answer :

To determine the coefficient of [tex]\( x^2 y \)[/tex] in the expansion of [tex]\( (8x + 7y)^3 \)[/tex], we can use the binomial theorem. The binomial theorem states:

[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]

Here, our variables are:
- [tex]\( a = 8x \)[/tex]
- [tex]\( b = 7y \)[/tex]
- [tex]\( n = 3 \)[/tex]

We need to find the term in the expansion that contains [tex]\( x^2 y \)[/tex]. This means we need the term where the power of [tex]\( x \)[/tex] is 2 and the power of [tex]\( y \)[/tex] is 1. This corresponds to finding the term where:
- [tex]\( x \)[/tex] appears to the power of 2
- [tex]\( y \)[/tex] appears to the power of 1

Let's denote the general term in the expansion by:
[tex]\[ \binom{n}{k} (8x)^{n-k} (7y)^k \][/tex]

We need to find the value of [tex]\( k \)[/tex] that satisfies our requirement. The exponent of [tex]\( x \)[/tex] in the term will be [tex]\( n-k \)[/tex], and the exponent of [tex]\( y \)[/tex] will be [tex]\( k \)[/tex]. We need:
[tex]\[ n - k = 2 \quad \text{and} \quad k = 1 \][/tex]

Since [tex]\( n = 3 \)[/tex], substituting these values:
[tex]\[ 3 - k = 2 \quad \implies \quad k = 1 \][/tex]

Now let's find the term when [tex]\( k = 1 \)[/tex]:
[tex]\[ \binom{3}{1} (8x)^{3-1} (7y)^1 \][/tex]

Simplify the term:
[tex]\[ \binom{3}{1} (8x)^2 (7y) = \binom{3}{1} (64x^2) (7y) \][/tex]

Next, calculate the binomial coefficient and multiply:
[tex]\[ \binom{3}{1} = 3 \][/tex]
[tex]\[ 3 \cdot 64x^2 \cdot 7y = 3 \cdot 64 \cdot 7 x^2 y = 1344 x^2 y \][/tex]

Hence, the coefficient of [tex]\( x^2 y \)[/tex] in the expansion of [tex]\( (8x + 7y)^3 \)[/tex] is:
[tex]\[ \boxed{1344} \][/tex]