1. ## Difference of Cubes

I have a real hard time understading the difference of cubes.
However, there are several questions on this topic in the coming June state exam.

How do I play with this topic?

QUESTIONS:

(1) Factor a^3b^3 - 8x^6y^9

(2) x^3 + 27

2. Originally Posted by symmetry
I have a real hard time understading the difference of cubes.
However, there are several questions on this topic in the coming June state exam.

How do I play with this topic?

QUESTIONS:

(1) Factor a^3b^3 - 8x^6y^9

(2) x^3 + 27
You have to memorize the factoring of the forms (x^3 +y^3) and (x^3 -y^3).
x^3 +y^3 = (x+y)(x^2 -xy +y^2) -----(i)
x^3 -y^3 = (x-y)(x^2 +xy +y^2) ------(ii)

(1) (a^3)(b^3) -8(x^6)(y^9)
= (ab)^3 -[2(x^2)(y^3)]^3
= [(ab) -2(x^2)(y^3)]*{(ab)^2 +(ab)[2(x^2)(y^3)] +[2(x^2)(y^3)]^2}
= [ab -2(x^2)(y^3)]*[(a^2)(b^2) +2ab(x^2)(y^3) +4(x^4)(y^6)] ----answer.

(2) x^3 +27
= x^3 +3^3
= (x+3)(x^2 -x*3x +3^2)

3. Originally Posted by symmetry
I have a real hard time understading the difference of cubes.
However, there are several questions on this topic in the coming June state exam.

How do I play with this topic?

QUESTIONS:

(1) Factor a^3b^3 - 8x^6y^9

(2) x^3 + 27
Ticbol says you have to memorise the factorisation of the sum and
difference of two cubes, but in fact if you know that sums or differences
of two cubes are involved you can deduce what the formulas are on the fly.

Consider f(x) = x^3+y^3, if x=-y, then f(x)=0, so (x+y) is a factor of f(x).

so we may write:

x^3+y^3=(x+y)(x^2 + A xy + y^2) = x^3 + A x^2y + xy^2 +yx^2 + Axy^2+y^3

Collect the terms with like powers of x and y:

x^3+y^3=(x+y)(x^2 + A xy + y^2) = x^3 + (A+1) x^2y + (A+1)yx^2 + y^3

So as the two sides are equal (A+1)=0, or A=-1, and:

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

This could have been obtained using polynomial long division once we know
that (x+y) is a factor of f(x).

The other factorisation can be obtained in a similar manner by observing
that (x-y) is a factor of x^3-y^3, or by replacing y by -y' throughout

RonL

4. Hello, symmetry!

I assume you know the sum and difference formulas.
. . $\displaystyle \begin{array}{cc}A^3 + B^3 \:=\:(A + B)(A^2 - AB + B^2) \\ A^3 - B^3 \:=\:(A - B)(A^2 + AB + B^2)\end{array}$

. . Understanding them?
. . Memorizing them?

. . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \underbrace{\text{plus or minus}}_{\downarrow}$
First, we must recognize the form: .$\displaystyle A^3 \pm B^3$
. . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \overbrace{\text{cube}}^{\uparrow}\;\overbrace{\te xt{cube}}^{\uparrow}$

The cubes break up into a linear factor and a quadratic factor.

. . . . . . $\displaystyle \underbrace{(A \;\pm \;B)}_{\text{linear}} \underbrace{(A^2 \;\mp \;AB \;+ \;B^2)}_{\text{quadratic}}$

To memorize the signs, think of the word SOAP.

. . $\displaystyle A^3 \;\pm \;B^3 \:=\:(A \;\pm\;B)\;(A^2 \;\mp \;AB \;+ \;B^2)$
. . . . . . . . . . . . . .$\displaystyle \uparrow\qquad\qquad\;\uparrow\qquad\quad\uparrow$
. . . . . . . . . . . . Same . . Opposite . Always Positive

5. Actually you don't need to memorize them, though memorization IS handy. As long as you recall that $\displaystyle a^3 \pm b^3$ is divisible by $\displaystyle a \pm b$ I think this provides a good exercise for that long division you mentioned in another thread.

-Dan

6. Originally Posted by topsquark
Actually you don't need to memorize them, though memorization IS handy. As long as you recall that $\displaystyle a^3 \pm b^3$ is divisible by $\displaystyle a \pm b$ I think this provides a good exercise for that long division you mentioned in another thread.

-Dan
There seems to be a bit of an echo in here today .

RonL

7. ## ok

I want to thank every tutor for your insight and tips.

Yes, my problem has been memorizing difference of cubes formulas.

8. Originally Posted by CaptainBlack
There seems to be a bit of an echo in here today .

RonL
Oops! I guess it pays to read the details of each post in the thread, which I obviously didn't do here.

-Dan