Polynomial question (1 Viewer)

[jxballistic]

Hi everybody,

For question a) I cannot figure out how to properly rearrange the question so that I can sub in numbers.
Ironically, I managed to do b)




The answer in the textbook is 9

Thanks!

 

[bleakarcher]

Let a=alpha, b=beta, c=gamma.
a^2(b+c)+b^2(a+c)+c^2(a+b)
=a^2[(a+b+c)-a]+b^2[(a+b+c)-b]+c^2[(a+b+c)-c]
=a^2(a+b+c)-a^3+b^2(a+b+c)-b^3+c^2(a+b+c)-c^3
=(a+b+c)[a^2+b^2+c^2]-(a^3+b^3+c^3)

Hopefully you can do it now.

Edit: lol'd at your 'lazycollegesenior' dp

 

[jxballistic]

Thanks
I lol'd the first time I saw it as well

I followed your method and got the same line as you in the end, but can you evaluate it? When I evaluated it I got -87

 

[Alkenes]

how come we have same DP 0.o lol

Thanks
I lol'd the first time I saw it as well

I followed your method and got the same line as you in the end, but can you evaluate it? When I evaluated it I got -87

 

[jxballistic]

how come we have same DP 0.o lol

Cos we're both cool

 

[Alkenes]

Lol ! True that!

cos we're both cool

 

[bleakarcher]

Let a=alpha, b=beta, c=gamma.
a^2(b+c)+b^2(a+c)+c^2(a+b)
=a^2[(a+b+c)-a]+b^2[(a+b+c)-b]+c^2[(a+b+c)-c]
=a^2(a+b+c)-a^3+b^2(a+b+c)-b^3+c^2(a+b+c)-c^3
=(a+b+c)[a^2+b^2+c^2]-(a^3+b^3+c^3)

a+b+c=-3
a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=(-3)^2-2(-2)=13
a^3+b^3+c^3=-3(a^2+b^2+c^2)+2(a+b+c)-3=-3(13)+2(-3)-3=-48
Hence, a^2(b+c)+b^2(a+c)+c^2(a+b)=(-3)(13)-(-48)=9

 

[jxballistic]

Let a=alpha, b=beta, c=gamma.
a^2(b+c)+b^2(a+c)+c^2(a+b)
=a^2[(a+b+c)-a]+b^2[(a+b+c)-b]+c^2[(a+b+c)-c]
=a^2(a+b+c)-a^3+b^2(a+b+c)-b^3+c^2(a+b+c)-c^3
=(a+b+c)[a^2+b^2+c^2]-(a^3+b^3+c^3)

a+b+c=-3
a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=(-3)^2-2(-2)=13
a^3+b^3+c^3=-3(a^2+b^2+c^2)+2(a+b+c)-3=-3(13)+2(-3)-3=-48
Hence, a^2(b+c)+b^2(a+c)+c^2(a+b)=(-3)(13)-(-48)=9

Wait so is
a^3 +b^3 + c^3 = -(B/A(a^2+b^2+c^2)+C/A(a+b+c) + 3D/A)
where A B C D are the coefficients of a cubic?

 

[bleakarcher]

^ looks right.

 

[jxballistic]

Are you sure there should be a negative outside the bracket? The one i copied in class didnt have a negative

 

[bleakarcher]

yep, the negative should be there

 

[jxballistic]

Ok yep the negative works in other questions. Thanks for your help