Simplifying Expressions

[GraciousV]

1 is the given expression and I am supposed to simplify 1 to (a-b)(b-c)(c-a)
1 ---ab^2+a^2c+bc^2-b^2c-c^2a-a^2b
2 ---a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)
I am now stucked at 2. Should I expand the binomial inside the parentheseses?

P/s: How do I write two of the expressions in the format below(LaTeX thing?)?
\(\displaystyle \sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}\) for \(\displaystyle i=1,\cdots,n-2\);

 

[Ishuda]

1 is the given expression and I am supposed to simplify 1 to \(\displaystyle (a-b)(b-c)(c-a)\)
1 ---\(\displaystyle a b^2 + a^2 c + b c^2 - b^2 c - c^2 a - a^2 b\)
2 ---\(\displaystyle a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)\)
I am now stucked at 2. Should I expand the binomial inside the parentheseses?

I don't know the answer to your question but I do notice that if we let
A = a - b
B = b - c
then
A + B = a - c
and you can write
(a-b)(b-c)(c-a) = - A B (A + B)

 

[GraciousV]

Should be simplified or either factorised. Because the required expression is simplified.

 

[Dale10101]

Maybe this

Should be simplified or either factorised. Because the required expression is simplified.

ab² + a²c + bc² - b²c - c²a - a²b
or
(ab² - b²c) + (a²c - c²a) + (bc² - a²b)
or
(b²)(a - c) + ac(a - c) + [(bc² - abc) + (abc - a²c)]
or
(b²)(a - c) + ac(a - c) + [(bc)(c - a) + (ab)(c - a)]
or
(b²)(a - c) + ac(a - c) + (bc)(c - a) + (ab)(c - a)
or
(-b²)(c - a) - ac(c - a) + (bc)(c - a) + (ab)(c - a)
or
(c - a)[-b² - ac + bc + ab]
or
(c - a)[(ab - ac) + (bc - b²)]
or
(c - a)[(a)(b - c) + (b)(c - b)]
or
(c - a)[(a)(b - c) - (b)(b - c)]
or
(c - a)[(b - c)(a - b)]
or
(c - a)(b - c)(a - b)
or
(a - b)(b - c)(c - a)

See if this works basically working backwards from (a - b)(b - c)(c - a)

 

[GraciousV]

Should be simplified or either factorised. Because the required expression is simplified.

ab² + a²c + bc² - b²c - c²a - a²b
or
(ab² - b²c) + (a²c - c²a) + (bc² - a²b)
or
(b²)(a - c) + ac(a - c) + [(bc² - abc) + (abc - a²c)]
or
(b²)(a - c) + ac(a - c) + [(bc)(c - a) + (ab)(c - a)]
or
(b²)(a - c) + ac(a - c) + (bc)(c - a) + (ab)(c - a)
or
(-b²)(c - a) - ac(c - a) + (bc)(c - a) + (ab)(c - a)
or
(c - a)[-b² - ac + bc + ab]
or
(c - a)[(ab - ac) + (bc - b²)]
or
(c - a)[(a)(b - c) + (b)(c - b)]
or
(c - a)[(a)(b - c) - (b)(b - c)]
or
(c - a)[(b - c)(a - b)]
or
(c - a)(b - c)(a - b)
or
(a - b)(b - c)(c - a)

See if this works basically working backwards from (a - b)(b - c)(c - a)

It seems like this solution required a lot of brainstorming!
Couldn't figure this out unless, I expand the required expression to the first expression.

 

[Dale10101]

Danke

BRILLIANT, Sir Dale !

Thanks Lord Denis.

 

[lookagain]

Alternative approach using steps similar to post # 5

ab² + a²c + bc² - b²c - c²a - a²b =

a²c - b²c - a²b + ab² - c²a + bc² =

c(a² - b²) - ab(a - b) - c²(a - b) =

c(a - b)(a + b) - ab(a - b) - c²(a - b) =

(a - b)[c(a + b) - ab - c²] =

(a - b)[ac + bc - ab - c²] =

(a - b)[bc - c² - ab + ac] =

(a - b)[c(b - c) - a(b - c)] =

(a - b)(b - c)(c - a)

 

[lookagain]

********square brackets not necessary...:roll:

They aren't necessary, just as the brackets aren't necessary in the 7th and the 11th steps
that you didn't bring up of Dale10101's post.

And the three places he used parentheses around b^2, and the one time around
-b^2, aren't necessary, either. And you didn't bring up those in Dale10101's post.

I did not question the use of parentheses in his 2nd line, because their addition
serves as an emphasis to separate three pairs of terms.

But, the brackets' use in mine is consistent in the flow of the steps between the
5th and the 8th lines, before changing over to the parentheses for the final step.

Therefore, it makes more sense to stay with them.

Apart from that, > > just as brilliant < < as Dale.

No, it is even more so than his.

I streamlined it with four fewer steps with greater efficiency built in but still used similar style
algebraic manipulation.

I avoided his lengthier, more complicated rewriting that he used in his 3rd line, for example.

(I'm glad he had a go-to method anyway at that point for versatility.)