Question

1)х²-2х+1= 2)р²-х²-2ху-у²= 3)а³-8+6а²-12а=

Answer 1

1.x^2-2x+1=(x-1)^2. 2. p^2-x^2-2xy-y^2=p^2-(x^2+2xy+y^2)=(p-x-y) (p+x+y) 3.a^3-8+6a^2-12a=(a-2) (a^2+8a+4)

Answer 2

1. To solve the equation x²-2x+1=0, we can use the fact that (x-1)² = x²-2x+1. So, we have:

x²-2x+1 = (x-1)² = 0

Therefore, the only solution is x=1.

2. To solve the equation р²-х²-2ху-у²=0, we can try to factor it by grouping the terms:

р²-х²-2ху-у² = (р²-х²) - 2ху - у² = (р-х)(р+х) - у² - 2ху = (р-х)(р+х-2у)

So, the solutions are:

р-х=0 => р=х р+х-2у=0 => y = (р+x)/2

Therefore, the solutions are (р, (р+x)/2) where р=x or р=-x.

3. To solve the equation а³-8+6а²-12а=0, we can factor it as follows:

а³+6а²-12а-8 = а²(а+6)-4(3а+2) = 0

So, the solutions are:

а²(а+6)=0 => а=0 or а=-6 3а+2=0 => а=-2/3

Therefore, the solutions are а=0, а=-6, or а=-2/3.