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

We have the equation3x^3 + 6x^2 - 4x +7 = 0with rootsa,b,c

we gotta find the equation with rootb+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) .............

Please Help!!