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 ++++++++
Hello,
I'll do it quite a rough way, I haven't thought of something simpler yet.
$\displaystyle \underbrace{(a+b+c)^3}_{(-2)^3=-8}=\underbrace{a^3+b^3+c^3}_{N}+{\color{red}3ab^2+ 3ac^2+3a^2b+3a^2c+3bc^2+3b^2c} \underbrace{+6abc}_{-6}$
$\displaystyle N=-2-{\color{red}\text{red part}}$
We'll take care of the red part.
Factor as much as possible :
$\displaystyle {\color{red}\text{red part}}=3 \bigg(ab(a+b)+ac(a+c)+bc(b+c)\bigg) $
Substitute from $\displaystyle a+b+c=-2$ :
$\displaystyle \begin{aligned} a+b&=&-2-c \\
a+c&=&-2-b \\
b+c&=&-2-a \end{aligned}$
Thus $\displaystyle {\color{red}\text{red part}}=3 \bigg(-2ab-abc-2ac-abc-2bc-abc\bigg)$
$\displaystyle {\color{red}\text{red part}}=3 \bigg(-2(ab+ac+bc)-3abc\bigg)=-6*7+9=-33$
Thus $\displaystyle N=-2-{\color{red}\text{red part}}=\boxed{31}$