Algebraic expression #4

[mathwannabe]

Hello everyone

Here is another one:

1) \(\displaystyle \dfrac{(a^2+b^2+c^2)^2-(a^2-b^2+c^2)^2}{4ab^2-4abc}=\)

\(\displaystyle =\dfrac{(a^2+b^2+c^2-a^2+b^2-c^2)(a^2+b^2+c^2+a^2-b^2+c^2)}{4ab^2-4abc}=\)

\(\displaystyle =\dfrac{2b^2(2a^2+2c^2)}{4ab(b-c)}=\)

\(\displaystyle =\dfrac{4b^2(a^2+c^2)}{4ab(b-c)}=\)

\(\displaystyle =\dfrac{b(a^2+c^2)}{a(b-c)}=\)

\(\displaystyle =\dfrac{a^2b+bc^2}{ab-bc}\)

The book says that the answer is \(\displaystyle \dfrac{a(b+c)}{b}\). How? What am I doing wrong?

 

[pka]

Hello everyone
Here is another one:
1) \(\displaystyle \dfrac{(a^2+b^2+c^2)^2-(a^2-b^2+c^2)^2}{4ab^2-4abc}=\)
\(\displaystyle =\dfrac{(a^2+b^2+c^2-a^2+b^2-c^2)(a^2+b^2+c^2+a^2-b^2+c^2)}{4ab^2-4abc}=\)
\(\displaystyle =\dfrac{2b^2(2a^2+2c^2)}{4ab(b-c)}=\)
\(\displaystyle =\dfrac{4b^2(a^2+c^2)}{4ab(b-c)}=\)
\(\displaystyle =\dfrac{b(a^2+c^2)}{a(b-c)}=\)
\(\displaystyle =\dfrac{a^2b+bc^2}{ab-bc}\)
The book says that the answer is \(\displaystyle \dfrac{a(b+c)}{b}\).

Once again, you simply don't understand basic operations.
\(\displaystyle (a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2\)
AND
\(\displaystyle (a^2-b^2+c^2)^2=a^4+b^4+c^4-2a^2b^2+2a^2c^2-2b^2c^2\)

 

[mathwannabe]

Once again, you simply don't understand basic operations.
\(\displaystyle (a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2\)
AND
\(\displaystyle (a^2-b^2+c^2)^2=a^4+b^4+c^4-2a^2b^2+2a^2c^2-2b^2c^2\)

So? I just went with difference of squares instead of squaring the trinomial... less cumbersome. Result is the same.

 

[mathwannabe]

\(\displaystyle \dfrac{(2^2 + 3^2 + 5^2)^2 - (2^2 - 3^2 + 5^2)^2}{(4 * 2 * 3^2)-(4*2*3*5)}\) \(\displaystyle = \dfrac{(4 + 9 + 25)^2 - (4 - 9 + 25)^2}{(8 * 9) -(8 * 15)} =\)

\(\displaystyle \dfrac{38^2 - 20^2}{72 - 120} = \dfrac{1444 - 400}{-48} = \dfrac{1044}{-48} = -21.75 < 0.\)

\(\displaystyle \dfrac{2(3 + 5)}{3} = \dfrac{16}{3} > 0.\)

The book is wrong again.

That book is EVIL! It tortures me x)

 

[Deleted member 4993]

Hello everyone

Here is another one:

1) \(\displaystyle \dfrac{(a^2+b^2+c^2)^2-(a^2-b^2+c^2)^2}{4ab^2-4abc}=\)

\(\displaystyle =\dfrac{(a^2+b^2+c^2-a^2+b^2-c^2)(a^2+b^2+c^2+a^2-b^2+c^2)}{4ab^2-4abc}=\)

\(\displaystyle =\dfrac{2b^2(2a^2+2c^2)}{4ab(b-c)}=\)

\(\displaystyle =\dfrac{4b^2(a^2+c^2)}{4ab(b-c)}=\)

\(\displaystyle =\dfrac{b(a^2+c^2)}{a(b-c)}=\)

\(\displaystyle =\dfrac{a^2b+bc^2}{ab-bc}\) <<< However this step is unnecessary and incorrect - the denominator should be (ab - ac)

The book says that the answer is \(\displaystyle \dfrac{a(b+c)}{b}\). How? What am I doing wrong?

.

 

[mathwannabe]

.

Yes, the denominator is \(\displaystyle ab-ac\) instead of \(\displaystyle ab-bc\). That was a typing mistake.

Thank you for pointing that out SK