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*-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
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) .............