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x2 + y4 + x4 +y2 = x6 + y6
unless you know what x and y are.
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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, then
x2 + y4 + x4 + y2 = 4, whilst
x6 + y6 = 2;
thus, they are two manifestly unequal quantities.
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How do you express x4 plus y4 if given x plus y equals a xy equals b?
(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
What is y to the 8th power as an product in four different was with only positive exponents?
y6 x y2 y4 x y4 y2 x y2 x y4 y2 x y2 x y2 x y2
Factor completely 32y4 -2x4 y4?
2y4(16 - x4) = 2y4 (4 + x2)(4 - x2) = 2y4(4 + x2)(2 + x)(2 - x)
X4 plus y4 divided by x plus y?
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
How do you solve y4-84 plus 5y2?
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.
