A challenging question (1 Viewer)

[QuaCk]

hmmmmm ok ....

The trick to this question is to know that:
a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc

Question:
If a+b+c=0, show that (2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3(2a-b)(2b-c)(2c-a)

Answer:
LHS = (2a-b)^3 + (2b-c)^3 + (2c-a)^3
= 7a^3 + 7b^3 + 7c^3 - 12ba^2 - 12cb^2 - 12ac^2 + 6ab^2 + 6bc^2 + 6ca^2 (after explanding and collecting like terms.
= 7(a^3 + b^3 + c^3) - 12ba^2 - 12cb^2 - 12ac^2 + 6ab^2 + 6bc^2 + 6ca^2
= 7[(a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc] - 12ba^2 - 12cb^2 - 12ac^2 + 6ab^2 + 6bc^2 + 6ca^2 (using the above statement)
= 21abc - 12ba^2 - 12cb^2 - 12ac^2 + 6ab^2 + 6bc^2 + 6ca^2 (since a+b+c=0)

RHS = 3(2a-b)(2b-c)(2c-a)
= 3(2a-b)(4bc-2ab+ac+c^2)
= 3(7abc - 4ba^2 - 4ac^2 - 4cb^2 + 2ca^2 + 2ab^2 + 2bc^2)
= 21abc - 12ba^2 - 12cb^2 - 12ac^2 + 6ab^2 + 6bc^2 + 6ca^2
= LHS

I hope that's right :|

 

[Trev]

that looks right but our teachers tell us to manipulate one side to look like the other, instead of changing both to make them equal because the hsc markers apparently don't like that...

 

[table for 1]

oh nuts. -.-" ... this is so gay. and i was thinking that you had to expand and then factorise to get the answer...omg...ok, that LHS=RHS just totally escaped my head...

well anyways, you should still be able to factorise the epanded LHS into the RHS, right? cos i can't...i got stuck. and, i don't even know the factorising of the sum of 3 cubes ! i only know the sum/difference of 2 cubes, and even then, i forgot it...ahhh....i'm starting to really panic right now. i'll be doing my hsc this yr, and i can't even solve maths questions from the beginning of the yr [and that's probably not a yr 12 Q, is it?]. excuse me while i go panic and cry. *utterly stressed*

 

[QuaCk]

as long as you can change both sides to an intermediary, it will still be accepted as being right.

 

[Trev]

as long as you can change both sides to an intermediary, it will still be accepted as being right.

can any1 confirm this, as we have otherwise been under a different impression.
we have been told we can expand the RHS but then just change the LHS to look like RHS
ie. (using QuaCk's working):
LHS = (2a-b)^3 + (2b-c)^3 + (2c-a)^3
= 7a^3 + 7b^3 + 7c^3 - 12ba^2 - 12cb^2 - 12ac^2 + 6ab^2 + 6bc^2 + 6ca^2 (after explanding and collecting like terms.
= 7(a^3 + b^3 + c^3) - 12ba^2 - 12cb^2 - 12ac^2 + 6ab^2 + 6bc^2 + 6ca^2
= 7[(a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc] - 12ba^2 - 12cb^2 - 12ac^2 + 6ab^2 + 6bc^2 + 6ca^2 (using the above statement)
= 21abc - 12ba^2 - 12cb^2 - 12ac^2 + 6ab^2 + 6bc^2 + 6ca^2 (since a+b+c=0)
= 3(7abc - 4ba^2 - 4ac^2 - 4cb^2 + 2ca^2 + 2ab^2 + 2bc^2)
= 3(2a-b)(4bc-2ab+ac+c^2)
= (2a-b)^3 + (2b-c)^3 + (2c-a)^3 = RHS
therefore LHS = RHS

 

[QuaCk]

i've been taught that there are 3 ways to prove these
either:

LHS = RHS
RHS = LHS
or
LHS = k = RHS

the last method is of course longest but in the situation that the function is very hard to factorise into the form required, leaving it expanded and comparing it against the expanded form of the opposite side will still be correct.
i'm sure of this.

 

[Templar]

can any1 confirm this, as we have otherwise been under a different impression.
we have been told we can expand the RHS but then just change the LHS to look like RHS

It is perfectly useable, just recap at the end LHS=RHS=k.

 

[Trev]

mmk, ive never heard of using 'k' with this stuff, ill ask my teacher bout it... thanx

 

[QuaCk]

it's not really 'k', it just represents a function of x different to what you are to prove.

 

[Slidey]

mmk, ive never heard of using 'k' with this stuff, ill ask my teacher bout it... thanx

There's no need to - he's correct.

Also if you can prove LHS-RHS=0, then that works, too.

 

[:: ck ::]

you're just gonna delete this after it gets answered eh?

i kno wot u mean by this

ppl have posted up q's in the 4unit forum... i spend like quite a bit of time tryin to answer it.. then when they see the answer... they just delete it....