Q: How do you factor x3 plus y3?

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x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)

x3 + x2 - 3x - 3 x(x2 + x - 3) - 3

(x + 7)(x2 - 7x + 49)

If that's 2x2, the answer is (x + 2)(x2 + 4)

x3-343

Related questions

x^2 - xy + y^2

It is an algebraic expression.

x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)

x3-y3

(x - y)(x^2 + xy + y^2

(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)

(x + y)3 + (x - y)3 = (x3 + 3x2y + 3xy2 + y3) + (x3 - 3x2y + 3xy2 - y3) = 2x3 + 6xy2 = 2x*(x2 + 3y2)

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.

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