(a) Find the binomial expansion of $(1-y)^6$ and simplify all terms. (b) Substitute $y = x - x^2$ into the expansion of $(1-y)^6$ and find the expansion of $(1 - (x - x^2))^6 = (1 - x + x^2)^6$ in powers of $x$ up to the term in $x^5$.

1. Problem Description

(a) Find the binomial expansion of $(1-y)^6$ and simplify all terms.

(b) Substitute $y = x - x^2$ into the expansion of $(1-y)^6$ and find the expansion of $(1 - (x - x^2))^6 = (1 - x + x^2)^6$ in powers of $x$ up to the term in $x^5$.

2. Solution Steps

(a) The binomial expansion of $(1-y)^6$ is given by:

$(1-y)^6 = \sum_{k=0}^6 \binom{6}{k} (1)^{6-k} (-y)^k$

Expanding the terms, we get:

$(1-y)^6 = \binom{6}{0} (1)^6 (-y)^0 + \binom{6}{1} (1)^5 (-y)^1 + \binom{6}{2} (1)^4 (-y)^2 + \binom{6}{3} (1)^3 (-y)^3 + \binom{6}{4} (1)^2 (-y)^4 + \binom{6}{5} (1)^1 (-y)^5 + \binom{6}{6} (1)^0 (-y)^6$

$(1-y)^6 = 1 - 6y + 15y^2 - 20y^3 + 15y^4 - 6y^5 + y^6$

(b) Substitute $y = x - x^2$ into the expansion of $(1-y)^6$:

$(1 - (x - x^2))^6 = 1 - 6(x - x^2) + 15(x - x^2)^2 - 20(x - x^2)^3 + 15(x - x^2)^4 - 6(x - x^2)^5 + (x - x^2)^6$

Now we expand each term and keep only the terms up to $x^5$:

$-6(x - x^2) = -6x + 6x^2$

$15(x - x^2)^2 = 15(x^2 - 2x^3 + x^4) = 15x^2 - 30x^3 + 15x^4$

$-20(x - x^2)^3 = -20(x^3 - 3x^4 + 3x^5 - x^6) = -20x^3 + 60x^4 - 60x^5 + 20x^6$. Keep up to $x^5$: $-20x^3 + 60x^4 - 60x^5$

$15(x - x^2)^4 = 15(x^4 - 4x^5 + 6x^6 - 4x^7 + x^8) = 15x^4 - 60x^5 + ...$. Keep up to $x^5$: $15x^4 - 60x^5$

$-6(x - x^2)^5 = -6(x^5 - 5x^6 + ...) = -6x^5 + ...$. Keep up to $x^5$: $-6x^5$

$(x - x^2)^6 = ...$. The lowest power of $x$ in this term is $x^6$, so we ignore it.

Now, we sum all terms up to $x^5$:

$1 + (-6x) + (6x^2 + 15x^2) + (-30x^3 - 20x^3) + (15x^4 + 60x^4 + 15x^4) + (-60x^5 - 60x^5 - 6x^5)$

$= 1 - 6x + 21x^2 - 50x^3 + 90x^4 - 126x^5$

3. Final Answer

(a) $(1-y)^6 = 1 - 6y + 15y^2 - 20y^3 + 15y^4 - 6y^5 + y^6$

(b) $(1 - x + x^2)^6 = 1 - 6x + 21x^2 - 50x^3 + 90x^4 - 126x^5 + ...$