Give an example of a sum of two cubes. Give an example of a difference of two cubes. Factor each.?

Question

25x cubed-8


25x cubed +8


If you want you can make up your owned problems I'm just trying to get an idea on how to do this. Thank you!

Answer 1

Your problems would be easier if you used 27 instead of 25. 27 is a perfect cube: 3³ = 27; 25 is not (it's a perfect square: 5² = 25).

The rules are:

a³ + b³ = (a + b)(a² - ab + b²)

and

a³ - b³ = (a - b)(a² + ab + b²)

So for your examples, a = 3x and b = 2:

27x³ - 8 = (3x - 2)(9x² + 6x + 4)

and

27x³ + 8 = (3x + 2)(9x² - 6x + 4)

Answer 2

x^3 - y^3 = (x - y)*(x^2 + x.y + y^2)

x^3 + y^3 = (x + y)*(x^2 - x.y + y^2)

These are called "algebraic factorisations", but of course they work for any particular values of x and y. For example, I know that 16^3 + 7^3 must have the factor 23, and that 22^3 - 5^3 must have the factor 17, without even working out the actual numbers.

Answer 3

the form of a sum/distinction of cubes: a^3 + b^3 = (cube root of each term) ("sq. the 1st" "unfavourable product' "sq. the final") a^3 + b^3 = (a + b) (? + ? + ?) "sq. the 1st" - sq. a, the 1st term interior the 1st set of brackets. a^3 + b^3 = (a + b) (a^2 + ? + ?) "unfavourable product" - take the made of the two words interior the 1st set of brackets, and then negate them (i.e. multiply the consequence by making use of unfavourable one). a circumstances b is ab, negated is-ab. a^3 + b^3 = (a + b) (a^2 - ab + ?) "sq. the final" - sq. the 2d term, b. a^3 + b^3 = (a + b) (a^2 - ab + b^2) back on your question. y^3 - 216 216 is recognizeable as 6^3 y^3 - 6^3 "cube root of each term. (y - 6)(? + ? + ?) "sq. the 1st" (y - 6)(y^2 + ? + ?) "unfavourable product" (y - 6)(y^2 + 6y + ?) "sq. the final" (y - 6)(y^2 + 6y + 36)

Answer 4

lol!


thanx!no problems thanks to God


good luck