# cubic formula

• Aug 23rd 2008, 02:16 AM
puneet
cubic formula
Q). Let a,b,c be the roots of x^3 + 2x^2 + 7x + 1 = 0.
Obtain A^3 +b^3 +C^3

SOLUTION).
a + b + c = -2
ab + bc + ca = 7
abc = -1
Now what is (a + b+ c)^3 = a^3 + b^3 + c^3 ++++++++
• Aug 23rd 2008, 02:44 AM
Moo
Hello,
Quote:

Originally Posted by puneet
Q). Let a,b,c be the roots of x^3 + 2x^2 + 7x + 1 = 0.
Obtain A^3 +b^3 +C^3

SOLUTION).
a + b + c = -2
ab + bc + ca = 7
abc = -1
Now what is (a + b+ c)^3 = a^3 + b^3 + c^3 ++++++++

I'll do it quite a rough way, I haven't thought of something simpler yet.

$\underbrace{(a+b+c)^3}_{(-2)^3=-8}=\underbrace{a^3+b^3+c^3}_{N}+{\color{red}3ab^2+ 3ac^2+3a^2b+3a^2c+3bc^2+3b^2c} \underbrace{+6abc}_{-6}$

$N=-2-{\color{red}\text{red part}}$

We'll take care of the red part.

Factor as much as possible :
${\color{red}\text{red part}}=3 \bigg(ab(a+b)+ac(a+c)+bc(b+c)\bigg)$

Substitute from $a+b+c=-2$ :

\begin{aligned} a+b&=&-2-c \\
a+c&=&-2-b \\
b+c&=&-2-a \end{aligned}

Thus ${\color{red}\text{red part}}=3 \bigg(-2ab-abc-2ac-abc-2bc-abc\bigg)$

${\color{red}\text{red part}}=3 \bigg(-2(ab+ac+bc)-3abc\bigg)=-6*7+9=-33$

Thus $N=-2-{\color{red}\text{red part}}=\boxed{31}$
• Aug 23rd 2008, 03:18 AM
puneet
solution
a^3 +b^3 +c^3=(a+b+c)^3+3abc-3(a+b+c)(ab+bc+ca)
=(-2)^3 +3(-1) -3(-2)(7)
=-8-3+42
=31 ans
Is this right
• Aug 23rd 2008, 03:22 AM
Moo
Quote:

Originally Posted by puneet
a^3 +b^3 +c^3=(a+b+c)^3+3abc-3(a+b+c)(ab+bc+ca)
=(-2)^3 +3(-1) -3(-2)(7)
=-8-3+42
=31 ans
Is this right

Yup (Tongueout)
Pardon me, I didn't know this factorisation and I made a mistake :)