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)
= (x+3)(x^2 -3x +9) -------------answer.
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
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 $
. . . . . . . . . . . . Same . . Opposite . Always Positive
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