20
(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.
If ( x = 16 ) and ( y = -16 ), then their combination ( x + y = 0 ) holds true. Thus, the values of ( x ) and ( y ) are 16 and -16, respectively.
y4.
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