Algebra question

A)

${(a-b)}^{3}= {a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3} = 1$

$1-3{a}^{2}b+3a{b}^{2}=1$

$3ab(b-a)=0$

$-3ab=0$

$ab=0$

B)

${(a+b)}^2={(a-b)}^{2}+4ab$

From previously, ab=0

$(a+b)^2={1}^{2}$

$a+b=\pm1$.

C)

We know a-b = 1, and a+b = 1 or a+b = -1.

Consider two cases:

Case 1:

$\begin{cases} a-b = 1\\ a+b=1 \end{cases}$

$2a = 2, a=1$

$\begin{cases} a=1 \\ b=0 \end{cases}$

Case 2:

$\begin{cases} a-b=1 \\ a+b = -1 \end{cases}$

$2a=0, a=0$

$\begin{cases} a=0 \\ b=-1 \end{cases}$

Our two solutions are $a, b = (1, 0)$ and $a, b = (0, -1)$.