One equation with two unknowns usually does not have a solution.
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.
the equation is A= y2-y2/x3-x2 after that you find the y-intercept by doing, b= y1+y2+y3-A(x1+x2+x3)/3
It is an algebraic expression.
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
One equation with two unknowns usually does not have a solution.
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.
x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)
the equation is A= y2-y2/x3-x2 after that you find the y-intercept by doing, b= y1+y2+y3-A(x1+x2+x3)/3
It is an algebraic expression.
3xyz
The required result will be 3xyz
(x2 - xy + y2)(x + y)
That's either the sum or difference of two cubes.
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
(x - y)(x^2 + xy + y^2
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