# Thread: Further maths. Cubics. What?

1. ## Further maths. Cubics. What?

Ok this 1 confused my maths teacher! You might need a pen and paper for this....

We have the equation 3x^3 + 6x^2 - 4x +7 = 0 with roots a,b,c
we gotta find the equation with root b+c, c+a, a+b
We did quite a lot of it this is how far we got...

written in the form x^3 - (sum)x^2 + (pairs)x - (product) = 0

sum = -6/3 = -2
pairs = -4/3
product = -7/3

New sum = (b+c) + (c+a) + (a+b)
= 2a + 2b + 2c
= 2(a+b+c)
=2*-2 = -4 We know this from the SUM worked out earlier

New Pairs = (b+c)(c+a) + (c+a)(a+b) + (b+c)(a+b)
= bc + ab + y^2 + ac + ac +bc + a^2 + ba + b^2 + ba + bc + ac
= a^2 + b^2 + c^2 + 3(cb + ca + ab)
= (a+b+c)^2 - 2(ab + ac + bc) + 3(-4/3) we know this from the PAIRS worked out earlier
=(-2)^2 - 2(-4/3) + (-4)
= 4 + 8/3 - 4
=8/3

Here comes the challenge...

New product = (b+c)(c+a)(a+b)
=(bc + ba + c^2 + ac)(a+b)
= bca + yb^2 + ba^2 + ab^2 + ac^2 + bc^2 + ca^2 + abc
= 2abc + a^2(b+c) + c^2(a+b) + b^2(c+a)
= 2(-7/3) .............

2. Originally Posted by djmccabie
Ok this 1 confused my maths teacher! You might need a pen and paper for this....

We have the equation 3x^3 + 6x^2 - 4x +7 = 0 with roots a,b,c
we gotta find the equation with root b+c, c+a, a+b
We did quite a lot of it this is how far we got...
It would be $[x-(b+c)][x - (a+c)] [x - (a+b)] = 0$.

Note, $(b+c)+(a+c)+(a+b) = 2(a+b+c) = -2\cdot \tfrac{6}{3} = -4$

And, $(b+c)(a+c)+(b+c)(a+b)+(a+c)(a+b) = (a^2+b^2+c^2) + 3(ab+ac+bc)$
This becomes $(a^2+b^2+c^2+2ab+2ac+2bc)+ab+ac+bc = (a+b+c)^2 + (ab+ac+bc) = (-\tfrac{6}{3})^2 - \tfrac{4}{3}$

Finally, $(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ac) - abc = (-\tfrac{6}{3})(-\tfrac{4}{3}) - (-\tfrac{7}{3})$

Use that to set up the cubic.

3. Thanks a lot!! you made a bit more work for yourself by repeating most of my working out though but thanks for the last part!!

4. Hello, djmccabie!

You did some awesomely excellent work!

Here's one more fact we can dig up . . .

We know: . $\begin{array}{cccc}a + b + c &=& \text{-}2 & {\color{blue}[1]}\\ \\[-4mm] ab+bc + ac &=& \text{-}\frac{4}{3} & {\color{blue}[2]}\\ \\[-4mm] abc &=& \frac{7}{3} & {\color{blue}[3]} \end{array}$

Square [1]: . $(a + b + c)^2 \;=\;(\text{-}2)^2 \quad\Rightarrow\quad a^2 + 2ab + 2ac + b^2 + 2bc + c^2\:=\:4$

$\text{We have: }\;a^2+b^2+c^2 + 2\underbrace{(ab + bc + ac)}_{\text{This is -}\frac{4}{3}} \:=\:4 \quad\Rightarrow\quad a^2 + b^2 + c^2 + 2\left(\text{-}\frac{4}{3}\right) \:=\:4$

Therefore: . $a^2+b^2+c^2\;=\;\frac{20}{3}$

5. just a quick check now. Do you get the answer 3x^3 + 12x^2 + 8x - 15 = 0 ?

(note coefficients must be integers)