Posted by **rachel** on Saturday, August 23, 2008 at 11:36am.

I'm not exactly sure how to factor cubed roots:

8x^3 - 1 = (2x)^3 - 1

that's what i get; can it be simplified anymore?

- math -
**rachel**, Saturday, August 23, 2008 at 11:43am
i'm looking at another one, and i don't know where to start:

x^3 - 2x^2 - 4x +8

hmm, cause u can't pull and x out, no quadratic . . .???

- math -
**Reiny**, Saturday, August 23, 2008 at 12:14pm
your first one fits the pattern for the difference of cubes

A^3 - B^3 = (A-B)(A^2 + AB + B^2)

so (2x)^3 - 1 = (2x-1)(4x^2 + 2x + 1)

test it by expanding my answer

for you second one, try grouping

I will get you started

x^3 - 2x^2 - 4x +8

= x^2(x-2) - 4(x-2)

= ......

- math -
**rachel**, Saturday, August 23, 2008 at 12:16pm
for function operations:

If f(x) = x^2 -4 and g(x) = [square root of] (2x+4)

g(f(a+2)) ---> then

x^2 - 4 +2

x^2 - x

[square root of] (2(x^2 - 2) +4)

= x [square root of] 2

wher does 'a' come from?, but is that right?

- math -
**Reiny**, Saturday, August 23, 2008 at 12:29pm
the x has been replaced by a+2

so since f(x) = x^2 -4 and g(x) = √ (2x+4)

g(f(x)) = √(2(x^2-4) + 4)

= √(2x^2 -4)

then g(f(a+2))

= √(2(a+2)^2 - 4)

=√(2(a^2 + 4a + 4) - 4)

= √(2a^2 + 8a + 4)

check my algebra, sometimes without writing it down first, I tend to make errors while just working it out only on the screen.

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