Here are the solutions.
1. Find the binomial expansion of x2(x−2y)3.
Step 1: Expand (x−2y)3 using the binomial theorem (a+b)n=∑k=0n(kn)an−kbk.
Here a=x, b=−2y, n=3.
(x−2y)3=(03)x3(−2y)0+(13)x2(−2y)1+(23)x1(−2y)2+(33)x0(−2y)3
=1⋅x3⋅1+3⋅x2⋅(−2y)+3⋅x⋅(4y2)+1⋅1⋅(−8y3)
=x3−6x2y+12xy2−8y3
Step 2: Multiply the result by x2.
x2(x3−6x2y+12xy2−8y3)=x5−6x4y+12x3y2−8x2y3
The final answer is x5−6x4y+12x3y2−8x2y3.
2. Find the term independent of x in the expansion of (x−x3)4.
Step 1: Write the general term Tk+1 for the expansion of (a+b)n.
Here a=x, b=−x3, n=4.
Tk+1=(k4)(x)4−k(−x3)k
=(k4)x4−k(−3)kx−k
=(k4)(−3)kx4−2k
Step 2: For the term independent of x, set the power of x to 0.
4−2k=0⟹2k=4⟹k=2
Step 3: Substitute k=2 into the general term.
T2+1=(24)(−3)2x4−2(2)
=6⋅9⋅x0
=54
The final answer is 54.
3. Find the first three terms in ascending powers of y in the expansion of (1−y)31.
Use the binomial expansion for fractional powers: (1+x)n=1+nx+2!n(n−1)x2+…
Here x=−y and n=31.
Step 1: First term.
1
Step 2: Second term.
n(−y)=31(−y)=−31y
Step 3: Third term.
2!n(n−1)(−y)2=231(31−1)y2
=231(−32)y2
=2−92y2
=−91y2
The first three terms are 1−31y−91y2.
4. Find the coefficient of x−3 in the expansion of ((x+1)(2x−x1)3).
Step 1: Expand (2x−x1)3.
Let a=2x and b=−x1.
(2x−x1)3=(03)(2x)3+(13)(2x)2(−x1)+(23)(2x)(−x1)2+(33)(−x1)3
=1⋅8x3+3⋅4x2⋅(−x1)+3⋅2x⋅(x21)+1⋅(−x31)
=8x3−12x+x6−x31
Step 2: Multiply the expansion by (x+1).
(x+1)(8x3−12x+x6−x31)
=x(8x3−12x+x6−x31)+1(8x3−12x+x6−x31)
=(8x4−12x2+6−x−2)+(8x3−12x+6x−1−x−3)
Step 3: Identify the term with x−3.
The term with x−3 is −x−3.
The coefficient of x−3 is −1.
The final answer is −1.