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Expanding cubic brackets (1 Viewer)
- Thread starter whitnall8
- Start date
pman
Banned
they won't ask you to expand any cubics that can't be done using pascals triangle ot binomial thereom
pman
Banned
but you won't get a cubic multinomial
- Joined
- Feb 16, 2005
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- 8,391
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- Male
- HSC
- 2006
Write it as (a + b + c + d)(a + b + c + d)(a + b + c + d)
If you multiply each of the same letter three times as per each factor you get the terms:
a³ + b³ + c³ + d³
If you multiply TWO of the same letter but one is different from each factor, notice that that there are always three (3!/2!) ways to do so depending on which letter from each factor you multiply in a certain order (e.g. the term a²b comes from a.a.b, a.b.a or b.a.a):
3a²b + 3a²c + 3a²d +
3b²a + 3b²c + 3b²d +
3c²a + 3c²b + 3c²d +
3d²a + 3d²b + 3d²c
If you multiply all letters to be different from each factor, also notice that that there are always six (3!) ways to do so depending on which letter from each factor you multiply in a certain order so you get the terms (e.g. the term abc comes from a.b.c, a.c.b, b.a.c, b.c.a, c.a.b and c.b.a):
6abc +
6abd +
6acd +
6bcd
Adding all the terms of up gives:
a³ + b³ + c³ + d³ + 3a²b + 3a²c + 3a²d + 3b²a + 3b²c + 3b²d + 3c²a + 3c²b + 3c²d + 3d²a + 3d²b + 3d²c + 6abc + 6abd + 6acd + 6bcd
If you multiply each of the same letter three times as per each factor you get the terms:
a³ + b³ + c³ + d³
If you multiply TWO of the same letter but one is different from each factor, notice that that there are always three (3!/2!) ways to do so depending on which letter from each factor you multiply in a certain order (e.g. the term a²b comes from a.a.b, a.b.a or b.a.a):
3a²b + 3a²c + 3a²d +
3b²a + 3b²c + 3b²d +
3c²a + 3c²b + 3c²d +
3d²a + 3d²b + 3d²c
If you multiply all letters to be different from each factor, also notice that that there are always six (3!) ways to do so depending on which letter from each factor you multiply in a certain order so you get the terms (e.g. the term abc comes from a.b.c, a.c.b, b.a.c, b.c.a, c.a.b and c.b.a):
6abc +
6abd +
6acd +
6bcd
Adding all the terms of up gives:
a³ + b³ + c³ + d³ + 3a²b + 3a²c + 3a²d + 3b²a + 3b²c + 3b²d + 3c²a + 3c²b + 3c²d + 3d²a + 3d²b + 3d²c + 6abc + 6abd + 6acd + 6bcd
This is going to end up pretty long regardless and Trebla's method is pretty good.
Alternatively, you can rewrite (a+b+c+d)^3 as [ (a+b) + (c+d) ] ^3
and then treat (a+b) and (c+d) as single terms.
So you'll end up with
(a+b)^3 + 3 (a+b)^2 (c+d) + 3(a+b) (c+d)^2 + (c+d)^3
Which you can then expand out again.
But regardless, a question like this will turn out to be very long and its unlikely to come out in an exam because of this and doesn't really test anything.
Alternatively, you can rewrite (a+b+c+d)^3 as [ (a+b) + (c+d) ] ^3
and then treat (a+b) and (c+d) as single terms.
So you'll end up with
(a+b)^3 + 3 (a+b)^2 (c+d) + 3(a+b) (c+d)^2 + (c+d)^3
Which you can then expand out again.
But regardless, a question like this will turn out to be very long and its unlikely to come out in an exam because of this and doesn't really test anything.
heteroskedastic
Member
- Joined
- Feb 24, 2010
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- 33
- Gender
- Male
- HSC
- 2011
just memorise some basic forms
(a + b = c)^3 - remember what it is
especially if you do ext2
(a + b = c)^3 - remember what it is
especially if you do ext2