Recall the binomial theorem.
[tex](a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k[/tex]
1. The binomial expansion of [tex]\left(1+\frac x3\right)^7[/tex] is
[tex]\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}[/tex]
Observe that
[tex]k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x[/tex]
[tex]k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2[/tex]
When we multiply these by [tex]8-9x[/tex],
• [tex]8[/tex] and [tex]\frac73 x^2[/tex] combine to make [tex]\frac{56}3 x^2[/tex]
• [tex]-9x[/tex] and [tex]\frac73 x[/tex] combine to make [tex]-\frac{63}3 x^2 = -21x^2[/tex]
and the sum of these terms is
[tex]\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}[/tex]
2. The binomial expansion is
[tex]\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k[/tex]
We get the [tex]a^6b^2[/tex] term when [tex]k=2[/tex] :
[tex]k=2 \implies \dbinom 82 2^{8-2\cdot2} a^{8-2} b^2 = 28 \cdot2^4 a^6 b^2 = \boxed{448} \, a^6b^2[/tex]
