Hello, I'm in algebra 2 and i wanted to know how to completely factor this: (a+b)^6 - (a-b)^6 . Thanks!!

Answer 1

Note that this is a difference of squares, so all you have to do is factor is as a difference of squares.

Reminder: x^2 - y^2 = (x-y)(x+y)

In this case, (a+b)^6 is the same as ((a+b)^3)^2

So (a+b)^6 - (a-b)^6 =

[ (a+b)^3 - (a-b)^3 ] [ (a+b)^3 + (a+b)^3 ]

Now we have sums and differences of cubes; the proper way to factor sums and differences of cubes are:

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

Verbally, the tactic would be: "square the first" "negative product" "square the last".

[ (a+b)^3 - (a-b)^3 ] [ (a+b)^3 + (a+b)^3 ] becomes
[ (a+b) - (a-b) ] [ (a+b)^2 + (a+b)(a-b) + (a-b)^2 ] [ (a+b) + (a-b) ] [ (a+b)^2 - (a+b)(a-b) + (a-b)^2 ]

simplifying, we get

(2b) [(a+b)^2 + (a+b)(a-b) + (a-b)^2] (2a) [ (a+b)^2 - (a+b)(a-b) + (a-b)^2 ]

(4ab) [a^2 + 2ab + b^2 + a^2 - b^2 + a^2 - 2ab + b^2] [ (a+b)^2 - (a+b)(a-b) + (a-b)^2 ]

(4ab) [3a^2 + b^2] [a^2 + 2ab + b^2 - (a^2 - b^2) + a^2 - 2ab + b^2]

(4ab) [3a^2 + b^2] [3b^2 + a^2]

This can't be factored any further.

Answer 2

you want to apply lengthy branch to divide x-6 into x^3-10x^2+19x+30. in case you're trying this outstanding, you receives x^2+4x-5. This elements into x+5 and x-a million, you get (x-6) x (x+5) x (x-a million) in case you do not recognize the thanks to do the polynomial lengthy branch or the thanks to component the binomial, ask a keep on with up question. i will briefly supply the easy idea of each and every: First, to do the lengthy branch, you spot that X-6 leads to 6 and the given polynomial leads to 30. 6 cases -5 is 30. so that you commence with '-5'. x-6 cases -5 is -5x+30. Subtracting you get X^3-10x^2+24x. -6 is going into 24x 4x cases. 4x cases x-6 is 4x^2-24x, subtracting supplies x^3-6x^2. -6 is going into 6^2 cases, providing you with a very last answer of x^2+4x-5. To component x^2+4x-5, purely see that 5 is 5 cases a million and that 4x is constructive. To get '-5' you'd be desiring to dissimilar a detrimental huge kind through a large, and because 4 is constructive, the better huge kind might want to be constructive (so their sum will be postive). therefore x+5 cases x-a million.

Answer 3

(a + b)^6 - (a - b)^6 =
((a + b)^3 + (a - b)^3)((a + b)^3 - (a - b)^3) =
4ab(a^2 + 3b^2)(3a^2 + b^2) =
4ab(a + ib√3)(a - ib√3)(b + ia√3)(b - ia√3)

Answer 4

(a+b)^6 - (a-b)^6
=[(a+b)^3 + (a-b)^3] x [(a+b)^3 - (a-b)^3]

note: x^y means x to the power of y

Answer 5

u probably know
1. x^2-y^2= (x+y)(x-y)
2. x^3-y^3= (x-y)(x^2+xy+y^2)
3. x^3+y^3= (x+y)(x^2-xy+y^2)
now firstly use 1
(a+b)^6-(a-b)^6=((a+b)^3+(a-b)^3)
((a+b)^3-(a-b)^3)
then use 2. & 3.
=(a+b+a-b)((a+b)^2-(a+b)(a-b)+(a-b)^2)
(a+b-(a-b))((a+b)^2+(a+b)(a-b)+(a-b)^2)
=(2a)(a^2+b^2+2ab-(a^2-b^2)+a^2+b^2-2ab)
(2b)(a^2+b^2+2ab+(a^2-b^2)+a^2+b^2-2ab)
=(2a)(a^2+3b^2)(2b)(3a^2+b^2)
=4ab(a^2+3b^2)(3a^2+b^2)
thanks & goodluck

Answer 6

(a+b)^6 - (a-b)^6
= [(a+b)^2 + (a+b)^2][(a+b)^4 - (a+b)^2(a-b)^2 + (a-b)^4]
= [(a^2+b^2+2ab) + (a^2+b^2-2ab][(a+b)^4 - (a^2-b^2) + (a-b)^4]
= [2(a^2+b^2)][(a+b)^4 - (a^2-b^2) + (a-b)^4]
= 2*(a+ib)*(a-ib)*[(a+b)^4 - (a^2-b^2) + (a-b)^4].............(x)

For (a+b)^4 - (a-b)^4
(a+b)^4 - (a-b)^4= ((a+b)^2)^2 - ((a-b)^2)^2
= (((a+b)^2)+((a-b)^2)^2) (((a+b)^2)-((a+b)^2))
= (((a^2+b^2+2ab)+((a^2+b^2-2ab)) (((a^2+b^2+2ab)-((a^2+b^2-2ab))
= (2*(a^2+b^2))*(4ab)

and for (a^2-b^2)
(a^2-b^2) = (a+b)(a-b)
Substituting in (x) we get
= 2*(a+ib)*(a-ib)*[ (2*(a^2+b^2)*(4ab) - (a+b)(a-b)]
= 2*(a+ib)*(a-ib)*[ (8*ab*(a^2+b^2) - (a+b)(a-b)]
= 2*(a+ib)*(a-ib)*[(8*ab*(a+ib)*(a-ib) - (a+b)(a-b)]

I hope that your answer

Answer 7

denote
x=a+b
y=a-b
xy=a^2-b^2
x+y=2a
x-y=2b

Therefore

x^6-y^6=(x^3)^2-(y^3)^2
=(x^3-y^3)(x^3+y^3)
=(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)
=2b(a^2+b^2+2ab+a^2-b^2 )(2a)(a^2+b^2+2ab - a^2+b^2+ a^2+b^2-2ab )
=2b(2a^2+2ab)(2a)(a^2+3b^2)

=8ab(a^2+ab)(a^2+3b^2)

Answer 8

well, you can take out the 6 as the factor:
6(a+b) - 6(a-b) = 6(a+b-(a-b))
= 6(a+b-a+b)
= 6(2b)
= 12b

or you can expand it
6(a+b) - 6(a-b)= 6a+6b-6a+6b
= 12b

Answer 9

12a^5b + 40a^3b^3 + 12ab^5

Answer 10

[{(a+b)^2}^3-{(a-b)^2}^3]
=[(a+b)^2-(a-b)^2]
[(a+b)^4+(a+b)^2(a-b)^2+(a-b)^4]
=4ab[(a+b)^2+(a+b)(a-b)+(a-b)^2]
[(a+b)^2-(a+b)(a-b)+(a-b)]^2]
=4ab[2a^2+2b^2+a^2-b^2]
[2a^2+2b^2-a^2+b^2)]
4ab[3a^2+b^2][a^2+3b^2]