Answer:
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To verify the identity \((a+b)^2 = a^2 + 2ab + b^2\) for the given values of \(a\) and \(b\), we will perform the calculations for each pair:
1. **\(a=8, b=5\)**
\[
(a+b)^2 = (8+5)^2 = 13^2 = 169
\]
\[
a^2 + 2ab + b^2 = 8^2 + 2(8)(5) + 5^2 = 64 + 80 + 25 = 169
\]
2. **\(a=6, b=8\)**
\[
(a+b)^2 = (6+8)^2 = 14^2 = 196
\]
\[
a^2 + 2ab + b^2 = 6^2 + 2(6)(8) + 8^2 = 36 + 96 + 64 = 196
\]
3. **\(a=7, b=6\)**
\[
(a+b)^2 = (7+6)^2 = 13^2 = 169
\]
\[
a^2 + 2ab + b^2 = 7^2 + 2(7)(6) + 6^2 = 49 + 84 + 36 = 169
\]
4. **\(a=9, b=4\)**
\[
(a+b)^2 = (9+4)^2 = 13^2 = 169
\]
\[
a^2 + 2ab + b^2 = 9^2 + 2(9)(4) + 4^2 = 81 + 72 + 16 = 169
\]
5. **\(a=10, b=8\)**
\[
(a+b)^2 = (10+8)^2 = 18^2 = 324
\]
\[
a^2 + 2ab + b^2 = 10^2 + 2(10)(8) + 8^2 = 100 + 160 + 64 = 324
\]
6. **\(a=9, b=8\)**
\[
(a+b)^2 = (9+8)^2 = 17^2 = 289
\]
\[
a^2 + 2ab + b^2 = 9^2 + 2(9)(8) + 8^2 = 81 + 144 + 64 = 289
\]
In all cases, we have verified that \((a+b)^2 = a^2 + 2ab + b^2\), confirming the identity is correct for the given values of \(a\) and \(b\).
