Question

Simplify :
\[=\frac{1}{4(b-a)^2}\left(b^3-3a^2b+6a^3\right)\]
NB: has a nice solution

Answer 1

should i try; partial fractions,
polynomial long division,
some other complicated method,
.../?

Answer 2

i am not confident in these methods

Answer 3

First guess is to use polynomial long division.

Answer 4

bother, polynomial long division is not my strong suite.

Answer 5

If I use long division I get\[{1 \over 4}\cdot\left({{4a^3 \over {(b-a)}^2}+2a+b}\right)\]Which is a little bit nicer.

Answer 6

the solution in the back of my book says
\[=\frac{2a+b}{4}\]

Answer 7

Well, \[{a^3 \over {(b-a)}^2} \neq 0\]so it looks like they left out a term to me.

Answer 8

yeah the solution could be incorrect,

Answer 9

\[=\frac{1}{4(b-a)^2}\left(b^3-3a^2b+6a^3\right)\]
\[=\frac{1}{4} \times (b-a)^2 \overline {\bigg)b^3-3a^2b+6a^3}\]
\[=\cdots \]

Answer 10

can you tell me the steps i always forget

Answer 11

First FOIL \((b-a)^2=b^2-2ab+a^2\). Then let's divide with respect to \(b\) first. so we need to find what we multiply \(b^2\) by to get \(b^3\). This is \(b\). So that's the first thing we put on top of the division symbol thing.

Answer 12

\[\quad\qquad\qquad\qquad b\]\[=\frac{1}{4} \times (b-a)^2 \overline {\bigg)b^3-3a^2b+6a^3}\]\[\quad\qquad\qquad\qquad b^3-2b^2a+a^2b\]

Answer 13

\[\quad\qquad\qquad\qquad b\]\[=\frac{1}{4} \times (b-a)^2 \overline {\bigg)b^3-3a^2b+6a^3}\]\[\quad\qquad\qquad\qquad b^3-2b^2a+a^2\]\[\quad\qquad\qquad\qquad \quad-3a^2b+2b^2a+5a^2\]

Answer 14

So now you have (with some steps not included) \[={1 \over 4}\left(b+ {2ab^2-4a^2b+6a^3 \over b^2-2ab+a^2}\right)\]Since \[b^3-3a^2b+6a^3 - (b^3-2b^2a+a^2b)=2ab^2-4a^2b+6a^3\]Remember you're subtracting an \(a^2b\) not just an \(a^2\)

Answer 15

\[\quad\qquad\qquad\qquad b-\frac{3}{2} a \]\[=\frac{1}{4} \times (b-a)^2 \overline {\bigg)b^3-3a^2b+6a^3}\]\[\quad\qquad\qquad\qquad b^3-2b^2a+a^2\]\[\quad\qquad\qquad\qquad \quad-3a^2b+2b^2a+5a^2\]\[\quad\qquad\qquad\qquad \quad \frac{3}{2}ab^2-3a^2b+\frac{3}{2}a^3\]

Answer 16

Now that we have this new equation, do long division one more time with respect to a this time. If I've done everything correct, it should be what we're looking for.