Question

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Answer 1

Recall the binomial theorem.

`(a+b)^n = \displaystyle \sum_(k=0)^n \binom nk a^(n-k) b^k`

1. The binomial expansion of
`\left(1+\frac x3\right)^7` is

`\left(1 + \frac x3\right)^7 = \displaystyle\sum_(k=0)^7 \binom 7k 1^(7-k) \left(\frac x3\right)^k = \sum_(k=0)^7 \binom 7k (x^k)/(3^k)`

Observe that

`k = 1 \implies \dbinom 71 \left(\frac x3\right)^1 = \frac73 x`

`k = 2 \implies \dbinom 72 \left(\frac x3\right)^2 = \frac73 x^2`

When we multiply these by
`8-9x`,

•
`8` and
`\frac73 x^2` combine to make
`\frac{56}3 x^2`

•
`-9x` and
`\frac73 x` combine to make
`-\frac{63}3 x^2 = -21x^2`

and the sum of these terms is

`\frac{56}3 x^2 - 21x^2 = \boxed{-\frac73 x^2}`

2. The binomial expansion is

`\left(2a - \frac 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`

We get the
`a^6b^2` term when
`k=2` :

`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`