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(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
Y3
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I understand this to be y3. Then the derivative is 3y2. If y is considered a so-called 'implicit function' of x then the derivative might be written 3y2 dy/dx.
y times y times y (or y3)
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The GCF of y3, y7, y8 is y3
y times y times y
In mathematics, the notation "y3" typically represents the cube of the variable "y," which is equivalent to y * y * y. This is a shorthand way of denoting the result of multiplying y by itself three times. The cube of a number is the number raised to the power of 3.
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
y3 x y3 - y (3)3 x 3(3) - 3 9 x 9 - 3 = ? 9 x 9= 81 81 - 3 = 78 I hope that solves your problem
y5 * y3 * y = y5 * y3 * y1 = y5+3+1 = y9
Y3
y(y - 2)(y + 2)
x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)
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