There is nothing to solve, unless it is given as an equation.
However, the above expression may be factorised, as
y4 - 18y2 + 81 = (y2 - 9)2 = (y - 3)2 (y + 3)2 .
Should you begin, however, with the equation,
y4 - 18y2 + 81 = 0,
then, a solution for y may be found in the following manner:
y4 - 18y2 + 81 = (y2 - 9)2 = (y - 3)2 (y + 3)2 = 0;
whence, y = ±3.
But note that this solution for y holds only when the trinomial expression with which we started is valued at zero. Otherwise, it could be any finite number.
Moral: we can factorise, or otherwise process, expressions; we can solve only equations.
x-7*y*4 >> solve it * is multiply-
If x = 3 and y = 4 then the answer is 2
The GCF is y4
4x-y4 what = 0
(x+y)4 = (x2+2xy+y2)2 = x4+4x3y+6x2y2+4xy3+y4
y4.
x-7*y*4 >> solve it * is multiply-
You cannot solve it since only one side of an equation is given. If the equation was y4 + 5y2 - 84 = 0 then y4 + 12y2 - 7y2 - 84 = 0 or y2*(y2 + 12) - 7*(y2 + 12) = 0 or (y2 - 7)*(y2 + 12) = 0 then y2 = 7 or y2 = - 12 y = +or- sqrt(7) and, if you are in the complex domain, also y = +or- i*sqrt(12) where i is the imaginary square root of -1.
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)
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.
(y^2 + 8)(y^2 + 2)
If x = 3 and y = 4 then the answer is 2
The GCF is y4
-3/4 if the question is 3x+4y
4 + 24 + 16 = 44
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
9y-81=9*(y-9)