find

maybe quadratic can help as it was in my C.P.P of quadratic eqns.

well one more thing...

if a + b + c = 0 then ax² + bx + c = 0 has a root 1 :P

\hspace{-16}$Let $\mathbf{a\;,b\;,c}$ be the roots of the equation $\mathbf{p(x)=x^3+px+q}$\\\\ Where $\mathbf{a+b+c=0\;,ab+bc+ca=p}$ and $\mathbf{abc=-q}$\\\\ Now $\mathbf{(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\Leftrightarrow a^2+b^2+c^2=-2(ab+bc+ca)}$\\\\ So $\mathbf{(a^2+b^2+c^2)=-2p\Leftrightarrow \boxed{ \bold{\sum a^2=-2p}}}$\\\\ Now $\mathbf{x^3+px+q=0\Leftrightarrow x^3=-\left(px+q\right)}$\\\\ Now $\mathbf{a\;,b}$ and $\mathbf{c}$ are the Roots of the equation.So it will satisfy it\\\\ $\mathbf{\sum a^3=-\left(p.\sum a+3q\right)=-\left(0+3q\right)=-3q\Leftrightarrow \boxed{\bold{\sum a^3=-3q}}}$\\\\ Where $\mathbf{\sum a=a+b+c}$ and $\mathbf{\sum a^3=a^3+b^3+c^3}$\\\\ Now $\mathbf{x^3=-(px+q)\Leftrightarrow x^6=p^2.x^2+q^2+2pqx}$\\\\ OR $\mathbf{x^7=p^2.x^3+q^2.x+2pq.x^2}$\\\\ Again $\mathbf{a\;,b}$ and $\mathbf{c}$ are the Roots of the equation.So it will satisfy it\\\\ So $\mathbf{\sum a^7=p^2.\sum a^3+q^2.\sum a+2pq.\sum a^2}$\\\\

\hspace{-16}\mathbf{\sum a^7=p^2.(-3q)+q^2.(0)+2pq.(-2p)=-7p^2q}$\\\\\\ $\boxed{\mathbf{\sum a^7=-7p^2q}}$\\\\\\ Where $\mathbf{\sum a^7=a^7+b^7+c^7}$ and $\mathbf{\sum a^3=a^3+b^3+c^3}$\\\\\\ So value of $\mathbf{7.\frac{\left(\sum a^2\right)^2.\sum a^3}{\sum a^7}}$\\\\\\ $\mathbf{7.\frac{(-2p)^2.(-3q)}{-7p^2q}=7.\frac{-12p^2q}{-7p^2q}=12}$\\\\\\ So $\mathbf{7.\frac{\left(\sum a^2\right)^2.\sum a^3}{\sum a^7}=12}$

Thank you sir...!!