1 yen
1 - y^(3)= 1^(3) - (3) = This factors to (1 - y)(1^(2) + y + y^(2)) More simply written as ( 1 - y)(1 + y + y^(2))
1
y-1/y2-y+1=y/y-1
x+y=1 x=1-y or y=1-x or if you fill that in 1-y+y=1 1 = 1 or 1-x+x = 1 1 = 1
(y * x) - y = y * (x - 1)
y = 3x + 1 y = 3(1) + 1 y = 3 + 1 Y = 4
(1-y)(1+y)
2x - y -(-1) = 2x - y + 1
2+y-1=0 i.e. y+1=0 Hence, y= -1.
y = 5x + 1 y - 1 = 5x (y - 1) / 5 = x
Let y = 3x y' = 3(x)' y' = 3(x1)' y' = 3[1x(1-1)] y' = 3(1x0) y' = 3(1 x 1) y' = 3 In general: y = xn y' nx(n-1)
4