(x2 + y2)(x + y)(x - y) = x4 - y4.
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.
√(40 x12y9) = 6.3246√(y)* x6 *y4 or 6.3246x6y4.5 ■
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
(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
18xy is the GCF of 72x2y and 18xy
(x+y)4 = (x2+2xy+y2)2 = x4+4x3y+6x2y2+4xy3+y4
There are infinitely many irrational numbers between any two numbers - rational or irrational.Suppose x and y are two irrational numbers.Consider x2 and y2. Is there any integer between them that is not a perfect square? If so, the square root of that number is an irrational between x and y.If not, consider x3 and y3 and look for an integer between them that is not a perfect cube. If there is then the cube root of that number will meet your requirements.If not, try x4 and y4 and then x5 and y5 etc. In a school exercise you are extremely unlikely to have to go as far as the cubes!
2y4(16 - x4) = 2y4 (4 + x2)(4 - x2) = 2y4(4 + x2)(2 + x)(2 - x)
The GCF is y4
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.