# Math Help - Factorisation

1. ## Factorisation

If a^2 + b^2 + c^2 = 13 and ab+bc+ca=6 then find the value of
a^3 + b^3 + c^3 -3abc

2. ## Re: Factorisation

The first two equations look remarkably like the terms of (a+b+c)^2 so first expand this item and you will find that with the information that you have you can show (a+b+c)^2 = 25.

Then expand (a+b+c)^3. Collecting the terms in the expanded form you will find that you can write it as:

$(a+b+c)^3=(a^2+b^2+c^2)(3a+3b+3c)-2(a^3+b^3+c^3 - 3abc)$

so

$25 \times (\pm 5) = 25 (3) (\pm 5) -2(a^3+b^3+c^3 - 3abc)$

so

$0 = 25 (2) (\pm 5) -2(a^3+b^3+c^3 - 3abc)$

$\pm 125= (a^3+b^3+c^3 - 3abc)$

There must be an easier way?

3. ## Re: Factorisation

Originally Posted by Kiwi_Dave
The first two equations look remarkably like the terms of (a+b+c)^2 so first expand this item and you will find that with the information that you have you can show (a+b+c)^2 = 25.

Then expand (a+b+c)^3. Collecting the terms in the expanded form you will find that you can write it as:

$(a+b+c)^3=(a^2+b^2+c^2)(3a+3b+3c)-2(a^3+b^3+c^3 - 3abc)$

so

$25 \times (\pm 5) = 25 (3) (\pm 5) -2(a^3+b^3+c^3 - 3abc)$

so

$0 = 25 (2) (\pm 5) -2(a^3+b^3+c^3 - 3abc)$

$\pm 125= (a^3+b^3+c^3 - 3abc)$

There must be an easier way?
I saw what you did about expansions, but I hate giving an answer that requires "seeing" something. I tried substitutions. It degenerates into a number of cases because you need to avoid dividing by zero and to address positive and negative roots. It was not pretty. Maybe I missed some neat simplification.

4. ## Re: Factorisation

Hello, Krishanu05!

My solution is similar to Kiwi-Dave's.

$\text{If }\,a^2 + b^2 + c^2 \,=\, 13\,\text{ and }\,ab+bc+ca\,=\,6$

$\text{then find the value of: }\:a^3 + b^3 + c^3 -3abc$

We have: . $\begin{Bmatrix}a^2+b^2+c^2 &=& 13 \\ 2ab + 2bc + 2ca &=& 12 \end{Bmatrix}$

Add: . $a^2+b^2+c^2+2ab + 2bc + 2ca \:=\:25 \quad\Rightarrow\quad (a+b+c)^2 \:=\:25$

Hence: . $a+b+c \:=\:5$

Cube: . $(a+b+c)^3 \:=\:5^3$

. . $a^3 + b^3 + c^3 + 3a^2b+3ab^2 + 3b^2c+3bc^2 + 3a^2c + 3ac^2 + 6abc \:=\:125$

. . $a^3 + b^3 + c^3 + 3a^2b+3ab^2 + 3b^2c+3bc^2 + 3a^2c + 3ac^2 + 9abc \:=\:125 + 3abc$

. . $(a^3+b^3+c^3) + (3a^2b+3ab^2 + 3abc) + 3b^2c+3bc^2 + 3abc) + (3a^2c + 3ac^2 + 3abc)$
. . . . . . . . . . . $=\:125 + 3abc$

. . $(a^3+b^3+c^3) + 3ab(a+b+c) + 3bc(a+b+c)+3ac(a+b+c) \:=\:125 + 3abc$

. . $(a^3+b^3+c^3) + 3\underbrace{(a+b+c)}_{\text{This is 5}}\underbrace{(ab + bc + ac)}_{\text{This is 6}} \;=\;125 + 3abc$

. . $(a^3+b^3+c^3) + 3(5)(6) \;=\;125 + 3abc$

Therefore: . $a^3+b^3+c^3 - 3abc \:=\:35$

5. ## Re: Factorisation

I can't fault Soroban's logic but I can't fault my own either so I will fill in the gaps of my attempt in the hopes that someone can point me to my error.

We agree to here:

$125=a^3 + b^3 + c^3 + 3a^2b+3ab^2 + 3b^2c+3bc^2 + 3a^2c + 3ac^2 + 6abc$

then I do something different:

$125=a^3 + b^3 + c^3 + a^2(3b+3c) +b^2(3a+3c)+c^2(3a+3b)+ 6abc$

So

$125=a^3 + b^3 + c^3 + a^2(3a+3b+3c) +b^2(3a+3b+3c)+c^2(3a+3b+3c)+ 6abc -3a^3 -3b^3 -3c^3$

$125=-2(a^3 + b^3 + c^3) + a^2(3a+3b+3c) +b^2(3a+3b+3c)+c^2(3a+3b+3c)+ 6abc$

or

$125=-2(a^3 + b^3 + c^3-3abc) + a^2(3 \times 5) +b^2(3 \times 5)+c^2(3 \times 5)$

or

$125=-2(a^3 + b^3 + c^3-3abc) + 3 \times 125$

or

$-2 \times 125=-2(a^3 + b^3 + c^3-3abc)$

finally

$125=(a^3 + b^3 + c^3-3abc)$

Now to take an entirely different tack. We have three variables and two constraint equations so there are an infinite number of solutions for a,b,c. Therefore, we should not be too quick to assume that there is only one answer to this problem.

6. ## Re: Factorisation

Originally Posted by Soroban
Hence: . $a+b+c \:=\:5$
Slight problem here. We have nothing to say that a, b, c are either positive or negative. So a + b + c might be -5 instead. The same problem is in Kiwi_Dave's solution.

Or am I missing something?

-Dan

7. ## Re: Factorisation

We could simply observe that a=0,b=2,c=3 is a solution with the answer 35
or that a=0,b=-2,c=-3 is a solution with the answer -35

Probably fits better with the Pre-university board too.