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

Question

i wanted to be clear on this. some people didn't understand my notation.

factor (a+b)^6 - (a-b)^6 here ^ i'm using for exponent. i didn't know how else to ask this so i'll post the question again. hopefully this clears it up. Thanks for the help!!!

Answer 1

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

Now, the previous answerers chose to factor the difference of 2 squares first, but I prefer to expand first.
= (a^6 + 6a^5 b + 15a^4 b² + 20a³b³ + 15a²b^4 + 6ab^5 + b^6) - (a^6 - 6a^5 b + 15a^4 b² - 20a³b³ + 15a²b^4 - 6ab^5 + b^6)

Now, we distribute the negative sign:
= a^6 + 6a^5 b + 15a^4 b² + 20a³b³ + 15a²b^4 + 6ab^5 + b^6 - a^6 + 6a^5 b - 15a^4 b² + 20a³b³ - 15a²b^4 + 6ab^5 - b^6

We combine like terms
= 12a^5 b + 40a³b³ + 12ab^5

We factor out 4ab
= 4ab(3a^4 + 10a²b² + 3b^4)

We factor out the biquadratic
= 4ab(3a² + b²)(a² + 3b²)

Therefore, the completely factored form is
(a + b)^6 - (a - b)^6 = 4ab(3a² + b²)(a² + 3b²)

^_^

Answer 2

you ought to use long branch to divide x-6 into x^3-10x^2+19x+30. in case you are trying this good, you will get x^2+4x-5. This factors into x+5 and x-a million, you get (x-6) x (x+5) x (x-a million) in case you don't be responsive to the thank you to do the polynomial long branch or the thank you to factor the binomial, ask a keep on with up question. i will temporarily provide the difficulty-free theory of each: First, to do the long branch, you spot that X-6 leads to six and the given polynomial leads to 30. 6 cases -5 is 30. so which you initiate off 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 an extremely final answer of x^2+4x-5. To factor x^2+4x-5, in basic terms see that 5 is 5 cases a million and that 4x is constructive. To get '-5' you will could multiple a unfavourable variety by making use of an outstanding, and because 4 is constructive, the greater advantageous variety could be constructive (so their sum would be postive). subsequently x+5 cases x-a million.

Answer 3

I try If I do not succeed completely I hope to help

first put A = (a+b)^3 = a^3 +3a^2 b+3ab^2 +b3

and B = (a-b)^3 = a^3 -3a^2b +3ab^2 -b3

your expression can be set like A^2 - B^2 = (A+B) (A-B)

(A+B) = 2 a^3 + 6 ab^2 and A-B = 6 a^2 b +2b^3

(A+B) = 4 ( 3a^5b + a^3b^3 + 9 a^3b^3 + 3ab^5)

result end 4 ab (3a^4 +10a^5b^5 +3b^4)

verify my answers

Answer 4

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

Let m = a + b, and n = a - b. This gives :

m^6 - n^6
= (m^3)^2 - (n^3)^2 (difference of 2 squares)
= (m^3 - n^3)(m^3 + n^3)
= (m - n)(m^2 + mn + n^2)(m + n)(m^2 - mn + n^2)

Now substitute back with a + b = m and a - b = n.
= (2b)(3a^2 + b^2)(2a)(a^2 + 3b^2)
= 4ab(3a^2 + b^2)(a^2 + 3b^2)

Answer 5

use the binomial theorem or a particularly neat way would be to take the difference of two squares, i.e,

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

Now it's already halfway resolved into factors, try the rest.

Hope this helps!

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+b)(a-b)]

Answer 7

(a+b)^6-(a-b)^6
=[(a+b)^3]^2-[(a-b)^3]^2
=[(a+b)^3+(a-b)^3]
[(a+b)^3-(a-b)^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{2(a^2+b^2)+(a^2-b^2)}]
[2b{2a^2+2b^2-(a^2-b^2)}]
=[2a(2a^2+2b^2+a^2-b^2)]
[2b(2a^2+2b^2-a^2+b^2)]
=[2a(3a^2-b^2]
[2b(3b^2+a^2)]]
=4ab(3a^2+b^2)(a^2+3b^2)
check for any typos