(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
y6 x y2 y4 x y4 y2 x y2 x y4 y2 x y2 x y2 x y2
x2 + y4 + x4 +y2 = x6 + y6unless you know what x and y are.* * * * *x2 + y4 + x4 + y2 ??I don't believe that this expression can be factorised or otherwise simplified.It certainly does not equal x6 + y6,for all x and all y:for example, if x = y = 1, thenx2 + y4 + x4 + y2 = 4, whilstx6 + y6 = 2;thus, they are two manifestly unequal quantities.
y4.
(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2
0
(x2 + y2)(x + y)(x - y) = x4 - y4.
It is x*x + y*y*y*y
X+2 Y=19 X-Y=4
7
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
y6 x y2 y4 x y4 y2 x y2 x y4 y2 x y2 x y2 x y2
x-7*y*4 >> solve it * is multiply-
If x = 3 and y = 4 then the answer is 2
y + y4 = y5 , possibly. Except that conventionally, the number (or coefficient) would be written first. y + y4 = y*(1 + y3) = y(1 + y)*(1 - y + y2)
4x-y4 what = 0
(x+y)4 = (x2+2xy+y2)2 = x4+4x3y+6x2y2+4xy3+y4