Need help with two Algebra 2 problems >_<

Hi, I was having a lot of trouble with these two algebra 2 problems.

You have to FACTOR each polynomial.

First one is:
(p+q)^3 - (p-q)^3

[^3 = to the power of 3]

and the second one is:

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

[^6 = to the power of 6 etc.]

PLEASE show it step by step! I really do appreciate it if you do help me.

Hi, I was having a lot of trouble with these two algebra 2 problems.

You have to FACTOR each polynomial.

First one is:
(p+q)^3 - (p-q)^3

[^3 = to the power of 3]

and the second one is:

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

[^6 = to the power of 6 etc.]

PLEASE show it step by step! I really do appreciate it if you do help me.

Lemme try the second one.

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

Let A = (a + b) ^ 3
Let B = (a - b) ^ 3

A^2 - B^2 = (A + B)(A - B)

Place A and B back in

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

Let A = (a + b)
Let B = (a - b)

(A^3 + B^3)(A^3 - B^3)

idk where to go from here, lol

Hi, I was having a lot of trouble with these two algebra 2 problems.

You have to FACTOR each polynomial.

First one is:
(p+q)^3 - (p-q)^3

[^3 = to the power of 3]

p3 + 3p2q + 3pq2 + q3 - (p3 -3p2q -3pq2 q3)
6p2q + 6pq2
6pq(p+q)

man... haven't done algebra since fist year uni...
fark been long long time.

Eh i got the first one..it factors down to 2q(3p^2 + 2q^2)

and Fiel I don't really know if that's right..at least I don't get it..do you think you can explain each step a little? :f6:

Lemme try the second one.

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

Let A = (a + b) ^ 3
Let B = (a - b) ^ 3

A^2 - B^2 = (A + B)(A - B)

Place A and B back in

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

Let A = (a + b)
Let B = (a - b)

(A^3 + B^3)(A^3 - B^3)

idk where to go from here, lol

from there, you use sums and differences of cubes.

(A^3 + B^3)(A^3 - B^3)

http://www.purplemath.com/modules/specfact2.htm

X^3+ Y^3 = (x + y)(X^2 - XY + Y^2)
X^3 -Y^3 = (X - Y)(X^2 + XY + Y^2)

argh I dont get it I looked in the back of the book and the answer is supposed to be 4ab(3a^2 + b^2)(a^2 + 3b^2)

but I dont understand how they got it!

Okay... here's basically what you need to know.

Difference of squares: m^2 - n^2 = (m+n)(m-n)
Difference of cubes: m^3 - n^3 = (m-n)(m^2+mn+n^2)
Sum of cubes: m^3 + n^3 = (m+n)(m^2-mn+n^2)

Basically what butterfli said. m and n can be anything. In the case of the first one m = p+q and n = p-q.

Can you do the second one from here or you need help?

using the sums and differences of cubes on (A^3 + B^3)(A^3 - B^3), you get

= (a+b)(a-b)(a²-ab+b²)(a²+ab+b²)

remember that fiel said

Let A = (a + b)
Let B = (a - b)

so that mess actually looks like

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

simplify it from there since its just a bunch of plus' and minus'. also because its plus' and minus', a buttload of things should cancel out.

 Answers!

Okay, you take Fiel's and butterfli's thing, and you have:

X [or A, which Fiel used] = a+b
Y [or B] = a-b

(X^3 + Y^3)(X^3 - Y^3)

X^3 + Y^3 = (X + Y)(X^2 - XY + Y^2)
X^3 - Y^3 = (X - Y)(X^2 + XY + Y^2)

And you combine them:

(X + Y)(X^2 - XY + Y^2)(X - Y)(X^2 + XY + Y^2)

Then you substitute X and Y for their original terms:

(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)

Simplify till the end.

2a(a^2+ab+b^2-(a^2-b^2)+a^2-ab+b^2)(a+b-a+b)(a^2+ab+b^2+a^2-b^2+a^2-ab+b^2)

2a(a^2+ab+b^2-a^2+b^2+a^2-ab+b^2)(a+b-a+b)(a^2+ab+b^2+a^2-b^2+a^2-ab+b^2)

2a(a^2+ab+b^2-a^2+b^2+a^2-ab+b^2)(a+b-a+b)(a^2+ab+b^2+a^2-b^2+a^2-ab+b^2

2a(a^2+3b^2)2b(3a^2+b^2)

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

Looks chaotic :f6:

Don't worry TS... precalc next year is easy.

Eww Precalc. Calculus is easier than Alg and PreCalc.

Eww Precalc. Calculus is easier than Alg and PreCalc.

Are you kidding me?
Boo, Calculus.

Integrals suck.

Eww Precalc. Calculus is easier than Alg and PreCalc.

Excluding vectors and some trig identities, the whole year in Precalc is basically review. I slept in that class and got an A.