How to factorise a^3-27b^3+7a^2b-21ab^2?
I tired using the identity (a+b) ^3 but got no where
How to factorise a^3-27b^3+7a^2b-21ab^2?
I used (a+b)^3 but couldn't get anywhere
Thanks in Advance
How to factorise a^3-27b^3+7a^2b-21ab^2?
I tired using the identity (a+b) ^3 but got no where
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)( ?????????)
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!!
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.
a^3-27b^3+7a^2b-21ab^2 = [a3 - (3b)3] + 7ab(a - 3b)
What did you get from here?