Question (1 Viewer)

[GaDaMIt]

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

Please show working

 

[acullen]

I'm not sure as to how they expect you to approach this problem, but I'll assume that you are to define two pronumerals as numeric constants.

As a+b+c=0

Let

a = 2
b = 3
.: 2+3+c=0
c=-5​

Proof

(2a-b)³+(2b-c)³+(2c-a)³ = 3(2a-b)(2b-c)(2c-a)

As a=2, b=3 and c=-5...

LHS
(2•2-3)³+(2•3-(-5))³+(2•(-5)-2)³
(4-3)³+(6+5)³+(-10-2)
1³+11³+(-12)³
1+1331-1728
-396

RHS
3(2•2-3)(2•3-(-5))(2•(-5)-2)
3(4-3)(6+5)(-10-2)
3(1)(11)(-12)
-396

.: LHS = RHS​

 

[GaDaMIt]

Are you sure you're allowed to just sub values in..?

 

[echelon4]

i'm wondering the same thing. Subbing in values would not prove the equation, it would only show that the equation would work with those numbers....

 

[+Po1ntDeXt3r+]

subbing in values is not a proof.. as it may not be true for all values

consider
(2a-b)=X
(2b-c)=Y
(2c-a)=Z

then hence X + Y+ Z = a + b + c = 0

and X^3+Y^3+Z^3 = 3XYZ

generally use the
(X + Y +Z) ^3 = 0 = X^3 + Y^3 + Z^3 + 6XYZ + 3 (X^2Y + X^2Z +Y^2Z + Y^2X + Z^2X + Z^2Y)

= 0 (from the X + Y+ Z = 0.. 0 cubed is zero)

now use the hint in the question
(2a-b)^3 + (2b – c)^3 + (2c – a)^3 = 3(2a-b)(2b-c)(2c-a)
which means ure trying to make
X^3+Y^3+Z^3 = 3XYZ

so add 3XYZ to ure equation
X^3 + Y^3 + Z^3 + 6XYZ + 3 (X^2Y + X^2Z +Y^2Z + Y^2X + Z^2X + Z^2Y) + 3XYZ - 3XYZ

factor out
gives u

X^3 + Y^3 + Z^3 + 3(X + Y + Z)(XY+YZ+ZX) - 3XYZ = 0

X^3 + Y^3 + Z^3 + 3(X + Y + Z)(XY+YZ+ZX) = 3XYZ

using X + Y+ Z = 0

X^3 + Y^3 + Z^3 = 3XYZ

then resubing the initial values

(2a-b)^3 + (2b – c)^3 + (2c – a)^3 = 3(2a-b)(2b-c)(2c-a)

Ok i havent done 3Unit maths in 2 yrs.. so im not sure if this is the shortest way

P.S took me 14 mins to refresh :S

 

[insert-username]

You have to prove the equation is true for all possible values of a, b, and c, not just a few random ones. I would assume that you expand (2a-b)³ + (2b – c)³ + (2c – a)³, then eliminate any instances of (a + b + c), as that equals 0, and that should equal the RHS. If I get time I'll expand it out and see if it works.

EDIT: Pointdexter's way is much better


I_F

 

[acullen]

Haha, what can I say, I got lazy.