Find the binomial expansion of x2(x−2y)3. 2. Find the term independent of x in the expansion of (x − 3/x)4.

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Here are the solutions to problems 1, 2, 3, and 4.

1. Find the binomial expansion of $x^2(x-2y)^3$ .

Expand $(x-2y)^3$ using the binomial theorem $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ .
Here, $a=x$ , $b=-2y$ , and $n=3$ .
$(x-2y)^3 = \binom{3}{0}x^3(-2y)^0 + \binom{3}{1}x^2(-2y)^1 + \binom{3}{2}x^1(-2y)^2 + \binom{3}{3}x^0(-2y)^3$ $= 1 \cdot x^3 \cdot 1 + 3 \cdot x^2 \cdot (-2y) + 3 \cdot x \cdot (4y^2) + 1 \cdot 1 \cdot (-8y^3)$ $= x^3 - 6x^2y + 12xy^2 - 8y^3$
Multiply the expanded expression by $x^2$ .
$x^2(x-2y)^3 = x^2(x^3 - 6x^2y + 12xy^2 - 8y^3)$ $= x^2 \cdot x^3 - x^2 \cdot 6x^2y + x^2 \cdot 12xy^2 - x^2 \cdot 8y^3$ $= x^5 - 6x^4y + 12x^3y^2 - 8x^2y^3$
The binomial expansion is $\boxed{x^5 - 6x^4y + 12x^3y^2 - 8x^2y^3}$ .

2. Find the term independent of $x$ in the expansion of $\left(x - \frac{3}{x}\right)^4$ .
Write the general term $T_{r+1}$ for the binomial expansion $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ .
Here, $a=x$ , $b=-\frac{3}{x} = -3x^{-1}$ , and $n=4$ .
$T_{r+1} = \binom{4}{r} (x)^{4-r} (-3x^{-1})^r$ $T_{r+1} = \binom{4}{r} x^{4-r} (-3)^r (x^{-1})^r$ $T_{r+1} = \binom{4}{r} (-3)^r x^{4-r-r}$ $T_{r+1} = \binom{4}{r} (-3)^r x^{4-2r}$
For the term independent of $x$ , the power of $x$ must be 0.
$4-2r = 0$ $2r = 4$ $r = 2$
Substitute $r=2$ back into the general term to find the term.
$T_{2+1} = T_3 = \binom{4}{2} (-3)^2 x^{4-2(2)}$ $T_3 = \frac{4!}{2!(4-2)!} (9) x^0$ $T_3 = \frac{4 \times 3}{2 \times 1} \times 9 \times 1$

Find the binomial expansion of x2(x-2y)3. 2. Find the term independent of x in the expansion of (x - (3)/(x))4.