Question

lf $\alpha ,\beta ,\gamma$ are the roots of $x^3+3x+3=0$, then ${\alpha }^5+{\beta }^5+{\gamma }^5=$

• A. $6$
• B. $18$
• C. $36$
• D. $45$

Answer 1

The correct option is D $45$

$x^3+0x^2+3x+3=0$
$x^3+a_1{x}^2+a_{2}x+3=0$
$s_1+a_1=0$
$s_1+a_1=0\Rightarrow s_1=0$(Since $a_1=0$)
$s_2+a_1{s}_1+{2a}_2=0\Rightarrow s_2+0+2.3=0\Rightarrow s_2=-6$
$s_3+a_1{s}_2+a_2{s}_1+{3a}_3=0$
$s_3+a_1{s}_2+a_2{s}_1+{3a}_3=0\Rightarrow s_3+0+3.0+3.3=0\Rightarrow s_3=-9$
$s_4+a_1{s}_3+a_2{s}_2+a_3{s}_1+4a_4=0\Rightarrow s_4+0+3.(-6)+0.3+0=0\Rightarrow s_4=18$ (Since $a_4=0$)
(Since $a_5=0$)
$s_5+a_1{s}_4+a_2{s}_3+a_3{s}_2+a_4{s}_1+5a_5=0\Rightarrow s_5+0+3.(-9)+3.(-6)+0+0=0\Rightarrow s_3=45$