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
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:
$\displaystyle (a+b+c)^3=(a^2+b^2+c^2)(3a+3b+3c)-2(a^3+b^3+c^3 - 3abc)$
so
$\displaystyle 25 \times (\pm 5) = 25 (3) (\pm 5) -2(a^3+b^3+c^3 - 3abc)$
so
$\displaystyle 0 = 25 (2) (\pm 5) -2(a^3+b^3+c^3 - 3abc)$
$\displaystyle \pm 125= (a^3+b^3+c^3 - 3abc)$
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.
Hello, Krishanu05!
My solution is similar to Kiwi-Dave's.
$\displaystyle \text{If }\,a^2 + b^2 + c^2 \,=\, 13\,\text{ and }\,ab+bc+ca\,=\,6$
$\displaystyle \text{then find the value of: }\:a^3 + b^3 + c^3 -3abc$
We have: .$\displaystyle \begin{Bmatrix}a^2+b^2+c^2 &=& 13 \\ 2ab + 2bc + 2ca &=& 12 \end{Bmatrix}$
Add: .$\displaystyle a^2+b^2+c^2+2ab + 2bc + 2ca \:=\:25 \quad\Rightarrow\quad (a+b+c)^2 \:=\:25$
Hence: .$\displaystyle a+b+c \:=\:5$
Cube: .$\displaystyle (a+b+c)^3 \:=\:5^3$
. . $\displaystyle a^3 + b^3 + c^3 + 3a^2b+3ab^2 + 3b^2c+3bc^2 + 3a^2c + 3ac^2 + 6abc \:=\:125$
. . $\displaystyle a^3 + b^3 + c^3 + 3a^2b+3ab^2 + 3b^2c+3bc^2 + 3a^2c + 3ac^2 + 9abc \:=\:125 + 3abc$
. . $\displaystyle (a^3+b^3+c^3) + (3a^2b+3ab^2 + 3abc) + 3b^2c+3bc^2 + 3abc) + (3a^2c + 3ac^2 + 3abc)$
. . . . . . . . . . . $\displaystyle =\:125 + 3abc$
. . $\displaystyle (a^3+b^3+c^3) + 3ab(a+b+c) + 3bc(a+b+c)+3ac(a+b+c) \:=\:125 + 3abc$
. . $\displaystyle (a^3+b^3+c^3) + 3\underbrace{(a+b+c)}_{\text{This is 5}}\underbrace{(ab + bc + ac)}_{\text{This is 6}} \;=\;125 + 3abc$
. . $\displaystyle (a^3+b^3+c^3) + 3(5)(6) \;=\;125 + 3abc$
Therefore: .$\displaystyle a^3+b^3+c^3 - 3abc \:=\:35$
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:
$\displaystyle 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:
$\displaystyle 125=a^3 + b^3 + c^3 + a^2(3b+3c) +b^2(3a+3c)+c^2(3a+3b)+ 6abc $
So
$\displaystyle 125=a^3 + b^3 + c^3 + a^2(3a+3b+3c) +b^2(3a+3b+3c)+c^2(3a+3b+3c)+ 6abc -3a^3 -3b^3 -3c^3$
$\displaystyle 125=-2(a^3 + b^3 + c^3) + a^2(3a+3b+3c) +b^2(3a+3b+3c)+c^2(3a+3b+3c)+ 6abc$
or
$\displaystyle 125=-2(a^3 + b^3 + c^3-3abc) + a^2(3 \times 5) +b^2(3 \times 5)+c^2(3 \times 5)$
or
$\displaystyle 125=-2(a^3 + b^3 + c^3-3abc) + 3 \times 125$
or
$\displaystyle -2 \times 125=-2(a^3 + b^3 + c^3-3abc)$
finally
$\displaystyle 125=(a^3 + b^3 + c^3-3abc)$
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.
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.