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How to factorise a^3-27b^3+7a^2b-21ab^2?
- Thread starter lPing7
- Start date
D
Deleted member 4993
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How to factorise a^3-27b^3+7a^2b-21ab^2?
I tired using the identity (a+b) ^3 but got no where
a^3-27b^3+7a^2b-21ab^2 = [a3 - (3b)3] + 7ab(a - 3b)
Now continue.....
How to factorise a^3-27b^3+7a^2b-21ab^2?
I used (a+b)^3 but couldn't get anywhere
Thanks in Advance
Taking a=a, b=-3b
(a+b) ^3= a^3 - 27b^3 - 9a^2b - 27b^2a
Subtracting result form original polynomial we get
-16a^b - 6 ab^2
This is as far as I can get
How to factorise a^3-27b^3+7a^2b-21ab^2?
I tired using the identity (a+b) ^3 but got no where
First recognize that 33 = 27 so the part of the expression can be written as a3 - (3 b)3 and you now have the difference of cubes. Also factor the the remaining part of the equation 7 a2 b - 21 a b2 on its own and see if that also helps. Continue from there or, if you are still stuck, show what you have done up to that point.
True that, and I'm not even sure whether to use a plus b whole cube or something else.After you expanded (a+b)^3, did you subtract the result
from a^3-27b^3+7a^2b-21ab^2 ? What did that leave?
I think that proper factorising means you'll be left
with a single expression, which will look like:
(a + b)( ?????????)
D
Deleted member 4993
Guest
a^3-27b^3+7a^2b-21ab^2 = [a3 - (3b)3] + 7ab(a - 3b)
What did you get from here?
What did you get from here?
True that, and I'm not even sure whether to use a plus b whole cube or something else.
WHY are you showing (a + b)^3; should be (a - 3b)^3; OK?
You have one wrong sign; should be:
(a - 3b)^3= a^3 - 27b^3 - 9a^2b + 27ab^2
Subtracting from original leaves: 16a^2b - 48ab^2
So we're now at:
(a - 3b)^3 + 16a^2b - 48ab^2
= (a - 3b)^3 + 16ab(a - 3b)
One more step: let's see your stuff!!
Oh my, I must have been blind to not realise that!
Anyways, \(\displaystyle (a-3b)^3 +16ab(a-3b) \)
\(\displaystyle (a-3b)((a-3b)^2 +16ab)\)
\(\displaystyle (a-3b)(a^2 +10ab+9b^2) \)
\(\displaystyle (a-3b)(a+9b)(a+b) \)
Looks alright to me, is it correct
Thanks for the guidance ?
Thank you Ishuda, that also got me to this same answer ?First recognize that 33 = 27 so the part of the expression can be written as a3 - (3 b)3 and you now have the difference of cubes. Also factor the the remaining part of the equation 7 a2 b - 21 a b2 on its own and see if that also helps. Continue from there or, if you are still stuck, show what you have done up to that point.
D
Deleted member 4993
Guest
a^3-27b^3+7a^2b-21ab^2 = [a3 - (3b)3] + 7ab(a - 3b)
What did you get from here?
Another way: (same as Ishuda's)
a^3-27b^3+7a^2b-21ab^2
= [a3 - (3b)3] + 7ab(a - 3b)
= [a - 3b](a2 + 3ab +9b2) + 7ab(a - 3b)
= (a-3b)(a2 + 3ab +9b2 + 7ab)
= (a-3b)(a2 + 10ab +9b2)
= (a-3b)(a + 9b)(a + b)
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