SOLUTION: Hello (40-something-old, returning to school -- very rusty, and a little embarrassed!) My problem: Factor 3x^6-192. Here's my approach.... Step 1: Factor out 3: 3(x^

Question 38041: Hello (40-something-old, returning to school -- very rusty, and a little embarrassed!)
My problem:
Factor 3x^6-192.
Here's my approach....
Step 1: Factor out 3:
3(x^6-64)
Step 2: Factor the "difference between two squares":
3(x^6-64) = 3(x^3-8)(x^3+8) Correct?
However, here's where I got stuck, so I decided to look at the answer key, and found this as the next step:
3(x^3-8)(x^3-8)(x^3+8)
Um, where did the "extra" (x^3-8) come from? Did I miss something really obvious here?...probably! I can't seem to think beyond this step, either. If you could remind me of the rule(s) involved, too, that's very helpful.
Thanks for your help!
Signed,
"Never-too-old to learn...I think"
Found 2 solutions by fractalier, askmemath:
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
The answer key is wrong. It happens every now and then...professors write these books and depend on overworked and underpaid graduate assistants to edit them...
You're okay so far...
3x^6 - 192 =
3(x^6 - 64) =
3(x^3 - 8)(x^3 + 8)
Now from here, there is a not-too-well-known way to factor the sum and difference of cubes...you'll have to memorize the method...but here it is...
a^3 + b^3 = (a + b)(a^2 - ab + b^2) and
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
so in our case
3(x^3 - 8)(x^3 + 8) =
3(x - 2)(x^2 + 2x + 4)(x + 2)(x^2 - 2x + 4)
and we're done...
And by the way, you should be embarassed if you stop wishing to learn, but proud of yourself for forging on ahead!
Answer by askmemath(368) (Show Source): You can put this solution on YOUR website!
Nope your work is perfect. Guess there must be a printing mistake in the book
What you are looking for is this
where and b = 8
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