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How to factirise this x3 y3 equation?
x3-y3
How do you factor x3 plus y3 divide x plus y?
x^2 - xy + y^2
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)
What is the answer to x plus y to 3rd power plus x minus y to 3rd power?
(x + y)3 + (x - y)3 = (x3 + 3x2y + 3xy2 + y3) + (x3 - 3x2y + 3xy2 - y3) = 2x3 + 6xy2 = 2x*(x2 + 3y2)
What is the solution to x3 plus 1331 equals y3?
One equation with two unknowns usually does not have a solution.
How to factirise this x3 y3 equation?
x3-y3
How do you factor x3 plus y3?
(x2 - xy + y2)(x + y)
How do you factor x3 plus y3 divide x plus y?
x^2 - xy + y^2
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)
What is the answer to x plus y to 3rd power plus x minus y to 3rd power?
(x + y)3 + (x - y)3 = (x3 + 3x2y + 3xy2 + y3) + (x3 - 3x2y + 3xy2 - y3) = 2x3 + 6xy2 = 2x*(x2 + 3y2)
What is the solution to x3 plus 1331 equals y3?
One equation with two unknowns usually does not have a solution.
When can a polynomial is said to be sum or difference of two cubes?
When it is of the form x3 + y3 or x3 - y3. x or y can have coefficients that are perfect cubes, or even ratios of perfect cubes eg x3 + (8/27)y3.
How do you factor x to the sixth minus y to the sixth?
x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)
Whats y3+x3+z3 equals to?
3xyz
x3+y3+z3=k?
The required result will be 3xyz
What is the coefficient of -y3 plus 7y3?
-y3 + 7y3
What is the answer of x2 plus y3 plus 15?
X2+Y3+15 All you can do is simplify it.
