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

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- Jan 19th 2007, 07:05 PMsymmetryDifference 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 - Jan 19th 2007, 07:37 PMticbol
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)

= (x+3)(x^2 -3x +9) -------------answer. - Jan 19th 2007, 11:49 PMCaptainBlack
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

the factorisation already obtained.

RonL - Jan 20th 2007, 03:41 AMSoroban
Hello, symmetry!

I assume you know the sum and difference formulas.

. . $\displaystyle \begin{array}{cc}A^3 + B^3 \:=\:(A + B)(A^2 - AB + B^2) \\ A^3 - B^3 \:=\:(A - B)(A^2 + AB + B^2)\end{array}$

Where is your difficulty?

. . 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 $

. . . . . . . . . . . .**S**ame . .**O**pposite .**A**lways**P**ositive

- Jan 20th 2007, 03:51 AMtopsquark
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 - Jan 20th 2007, 04:12 AMCaptainBlack
- Jan 20th 2007, 04:53 AMsymmetryok
I want to thank every tutor for your insight and tips.

Yes, my problem has been memorizing difference of cubes formulas. - Jan 21st 2007, 04:21 AMtopsquark