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ThePerfectHacker · #2 October 3rd, 2008, 07:02 AM

Quote:

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.

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djmccabie
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