Bash me up please!

$p=a+b+c=1 \\ q=ab+bc+ac \\ r=abc$

$p^2=1=\sum a^2+2(ab+bc+ac) \implies 2+2q=1 \implies q=-\dfrac{1}{2}$

$f(x)=(x-a)(x-b)(x-c)=x^3-px^2+qx-r=x^3-x^2-\dfrac{x}{2}-r \quad \quad f(a)=f(b)=f(c)=0$

$f(a)=0 \implies a^3-a^2-\dfrac{a}{2}-r=0 \implies a^3=a^2+\dfrac{a}{2}+r \\ \sum a^3=3=\sum a^2+\dfrac{\sum a}{2}+3r=\dfrac{5}{2}+3r \implies 3r=\dfrac{1}{2} \implies r=\dfrac{1}{6}$

$a^3=a^2+\dfrac{a}{2}+\dfrac{1}{6} \implies a^4=a^3+\dfrac{a^2}{2}+\dfrac{a}{6} \\ \sum a^4=\sum a^3+\dfrac{\sum a^2}{2}+\dfrac{\sum{a}}{6}=3+\dfrac{2}{2}+\dfrac{1}{6}=\dfrac{25}{6}$

$a^3=a^2+\dfrac{a}{2}+\dfrac{1}{6} \implies a^5=a^4+\dfrac{a^3}{2}+\dfrac{a^2}{6} \\ \sum a^5=\sum a^4+\dfrac{\sum a^3}{2}+\dfrac{\sum a^2}{6}=\dfrac{25}{6}+\dfrac{3}{2}+\dfrac{2}{6}=\dfrac{36}{6}=6$

$a^5+b^5+c^5=\color{#20A900}{\boxed{\boxed{6}}}$

Using Newton sum.

We get,
$\Rightarrow ab+bc+ca=-\dfrac{1}{2}$

$\Rightarrow abc=\dfrac{1}{6}$

$\Rightarrow a^4+b^4+c^4=\dfrac{25}{6}$

And,

$a^5+b^5+c^5=(a+b+c)(a^4+b^4+c^4)-(ab+bc+ca)(a^3+b^3+c^3)+abc(a^2+b^2+c^2)$

$a^5+b^5+c^5=\dfrac{25}{6}+\dfrac{3}{2}+\dfrac{1}{3}=\dfrac{36}{6}$

$\therefore a^5+b^5+c^5=\boxed{6}$