how do you factor these problems?

Question

d^6 + h^3

3c^4 - 81c

x^3 - x

I dont understand these?

the book says that they should end up being the sum or difference of 2 cubes:

x^3 + y^3 = (x+y)(x^2 - xy +y^2)
x^3 - y^3 = (x-y)(x^2 +xy +y^2)

can someone show me how to do these step by step?

Answer 1

x^3 - x = x(x^2 - 1) = (x) (x+1) (x-1) - this is difference of two squares.

3c^4 - 81c = we can factor a 3c from the the term, getting:

3c(c^3-27) Now, for the inner part you can use the (x-y) term, and finish with:

(3c) (c-3) (c^2 +3c +9)

Top one:

X^3 +Y^3 is d^6+h^3 So X is d^2, and Y is h

(d^2+h) (d^4 -d^2h + h^2)

Factor out whatever lower terms you can first, to get it into a difference of two squares, or difference of two cubes, then follow the rules.

Answer 2

d^6 + h^3 = (d^2)^3 + h^3
from hereon you can use the formula for the sum of 2 cubes.

3c^4 - 81c = 3c*(c^3 - 9) = 3c*(c^3 - 3^3)
from hereon you can use the formula for the diference of 2 cubes for the terms in the parenthesis

x^3 - x = x*(x^2 - 1)
ok something is wrong here. Either the equation is x^4 - x or you are supposed to use the formula for the difference between 2 squares.

Answer 3

ok these are great fun. first of all your formula for getting the answer is F^3 + S^3 = (F+S)(F^2 - FS + S^2) or
F^3 - S^3 = (F- S)(F^2+ FS + S^2)

F= the first number
S= the second number

so #1, d^6 + h^3 = (d^2)^3 + (h)^3

F= d^2 and S= h^3

(d^2 + h^3)(d^4 - d^2h^3 + h^6) is your answer

#2, 3c^4 - 81c= 3c(c^3 - 27) = 3c(c^3 - 3^3)= 3c[ (c)^3 - (3)^3]

F=c
S=3 note: after factoring F and S, you must multiply the
answer by 3c

3c(c-3)(c^2 + 3c + 9) is your answer

plz vote mine best

Answer 4

d^6 + h^3 = (d^2)^3 + h^3 = (d + h)(d^2 - dh + h^2)

3c^4 - 81c = 3c (c^3 - 27) = 3c (c^3 - 3^3) = 3c (c - 3)(c^2 + 3c + 9)

x^3 - x = x (x^2 - 1) = x (x +1)(x - 1)

Lat one can't be written in the form of a^3 - b^3 unless we are ready to take Cube root of x as one term .

Answer 5

d^6 + h^3 These terms have nothing in common with each other and therefore has no factors that I can see.

3c^4 - 81c
3c(c^3-21)

x^3 - x
x(x^2-1)

Answer 6

d^6 + h^3 =
(d^2 + h)(d^4 -d^2h + h^2)

3c^4 - 81c =
3c(c^3 - 27) =
3c(c - 3)(c^2 +3c + 9)


x^3 - x =
x(x^2 - 1) =
x(x - 1)(x + 1)

Answer 7

you need to get help from your school. getting your homework done here is cheating, and it won't help you, as you'll never learn. Get help from your school.