Proving this identity? Help pleaseeee :) (1 Viewer)

[Prawnchip]

(a + b + c)(ab + bc + ca) - abc = (a + b)(b + c)(c + a)

Can anyone show the full working out? My algebra stinks.
After working it out, I got up to here, now what?
LHS = ab( a + b) + ac (a + c) + bc (b + c) + 2abc

 

[Verify]

Is this from Cambridge?

 

[gosfa747]

looks familiar Cambridge 3 unit 1b 14a

 

[Prawnchip]

Lul yes its from cambridge

 

[superSAIyan2]

M1 Expand and collect like terms on both sides

M2 let a,b and c be the roots of the cubic eqn fx^3 + gx^2 + hx + j = 0

LHS = sum of roots x sum doubles - product = - (gh/f^2 + j/f)

SINCE a+b+ c = -g/f therefore a+b = -g/f -c

RHS = (-g/f -a)(-g/f -b)(-g/f - c)
= - (g/f+a)(g/f+ b)( g/f + c)
= - (g^3/f^3 + g^2/f^2(sum of roots) + g/f (sum of doubles of roots) + product of roots)
= - (g^3/f^3 - g^3/f^3 + gh/f^2 + j/f)
= LHS

 

[cricketfan1997]

M1 Expand and collect like terms on both sides

M2 let a,b and c be the roots of the cubic eqn fx^3 + gx^2 + hx + j = 0

LHS = sum of roots x sum doubles - product = - (gh/f^2 + j/f)

SINCE a+b+ c = -g/f therefore a+b = -g/f -c

RHS = (-g/f -a)(-g/f -b)(-g/f - c)
= - (g/f+a)(g/f+ b)( g/f + c)
= - (g^3/f^3 + g^2/f^2(sum of roots) + g/f (sum of doubles of roots) + product of roots)
= - (g^3/f^3 - g^3/f^3 + gh/f^2 + j/f)
= LHS

woah serious?

 

[Drongoski]

Another method - not necessarily better.


P(a) = (a+b+c)(ab+bc+ca) - abc is a polynomial in 'a'.

Now P(-b) = (-b+b+c)(ab +bc -bc) - abc = 0

.: (a- [-b]) = (a+b) is a factor of the LHS

Similarly, (b+c) and (c+a) are factors of the LHS

.: LHS = k(a+b)(b+c)(c+a) for some constant k

Since coeff of term in, say, a²b of the LHS is 1, k=1

.: LHS = (a+b)(b+c)(c+a)

QED

 

[seanieg89]

Another method - not necessarily better.


P(a) = (a+b+c)(ab+bc+ca) - abc is a polynomial in 'a'.

Now P(-b) = (-b+b+c)(ab +bc -bc) - abc = 0

.: (a- [-b]) = (a+b) is a factor of the LHS

Similarly, (b+c) and (c+a) are factors of the LHS

.: LHS = k(a+b)(b+c)(c+a) for some constant k

Since coeff of term in, say, a²b of the LHS is 1, k=1

.: LHS = (a+b)(b+c)(c+a)

QED

To add to this more sophisticated method:

1. The "similarly" follows from the fact that the multivariate polynomial on the LHS of this equation is symmetric: that is, interchanging any two of a,b,c does not change it as a polynomial.

2. The LHS must be of the form k(a+b)(a+c)(b+c) after establishing those three factors of the LHS, as the LHS is of degree 3. If we did not know this then k would have to be replaced with something potentially more complex.

Exploiting symmetry, using the factor theorem, and using degree are together a powerful way of dealing with multivariable polynomials in general, but this isn't so important for the sort of questions they can ask you in the HSC.

 

[Verify]

Wow I don't remember the answer being this sophisticated.

 

[seanieg89]

Wow I don't remember the answer being this sophisticated.

It doesn't have to be, the earlier solutions posted work fine. Drogonski's method is an illustration of a more powerful tool.

 

[Drongoski]

It doesn't have to be, the earlier solutions posted work fine. Drogonski's method is an illustration of a more powerful tool.

That's true. One could more easily have fully expanded both the LHS and the RHS and shown them to be identical.

 

[Carrotsticks]

 

[Prawnchip]

Thank you for the replies!