Yes, of course, you can do that. Further, x^3- y^3= (x- y)(x^2 + xy+ y^2) and x^3+ y^3= (x+ y)(x^2- xy+ y^2) so that can be written as (x- y)(x+ y)(x^2+ xy+ y^2)(x^2- xy+ y^2).RTCNTC said:Factor x^6 - y^6 as a difference of cubes.
Solution:
(x^3 - y^3)(x^3 + y^3)
I now proceed by applying the difference and sum of cubes accordingly, right?
The formula for "X^6 - y^6 As Difference of Cubes" is (X^3 - y^3)(X^3 + y^3). It follows the pattern of (a^3 - b^3) = (a - b)(a^2 + ab + b^2).
The main difference between "X^6 - y^6 As Difference of Cubes" and "X^6 + y^6 As Sum of Cubes" is the sign between the two terms. In "X^6 - y^6 As Difference of Cubes", the sign is negative while in "X^6 + y^6 As Sum of Cubes", the sign is positive.
Yes, "X^6 - y^6 As Difference of Cubes" can be factored further into (X - y)(X + y)(X^2 + xy + y^2)(X^2 - xy + y^2). This is known as the difference of squares rule.
"X^6 - y^6 As Difference of Cubes" is a special case of the difference of powers formula and is used in many mathematical concepts such as algebraic expressions, polynomial functions, and binomial expansions. It is also helpful in solving complex equations and simplifying expressions.
Yes, the concept of "X^6 - y^6 As Difference of Cubes" can be applied to any powers as long as the powers are integers. For example, (X^4 - y^4) can be factored as (X - y)(X + y)(X^2 + y^2).