Fun with Ramanujan's number 1729.

By Vieta's formula , we have $\alpha + \beta+\gamma+\delta = -7$ and $\alpha\beta + \alpha \gamma + \alpha \delta + \beta\gamma+\beta \delta + \gamma\delta = 2$ . Since

$\begin{aligned} \alpha^2 + \beta^2+\gamma^2+\delta^2 & = (\alpha + \beta+\gamma+\delta)^2 - 2(\alpha\beta + \alpha \gamma + \alpha \delta + \beta\gamma+\beta \delta + \gamma\delta) \\ & = (-7)^2 - 2(2) = \boxed{45} \end{aligned}$

The given equation can be rewritten as $x^4+7x^3+2x^2-1720=0$ . So $α+β+\gamma+\delta=-7, αβ+α\gamma+α\delta+β\gamma+β\delta+\gamma\delta=2$ . Therefore $α^2+β^2+\gamma^2+\delta^2=(α+β+\gamma+\delta) ^2-2(αβ+α\gamma+α\delta+β\gamma+β\delta+\gamma\delta)=(-7)^2-2\times 2=49-4=\boxed {45}$ . The equation can be generalized as $1\times x^n+7\times x^{n-1}+2\times x^{n-2}+9=1729$ .

If $r_1, r_2,r_3$ and $r_4$ are the roots of the equation then by multinomial theorem $(r_1+r_2+r_3+r_4)^2=\sum_{r_1+r_2+r_3+r_4=2} { 2 \choose k_1,k_2,k_3,k_4} \prod_{t=1}^{4} r_t^{k_t}=\sum_{k_1+\cdots k_4=2}\frac{n!}{k_1! k_2!\cdots k_4!}\prod_{t=1}^4r_t^{k_t} \\ ={ 2 \choose 2 ,0,0,0}r_1^{2}+{2 \choose 1,1,0,0} r_1 r_2 +{2\choose 0,2,0,0} r_2^2 + {2\choose 0,0,0,2} r_4^2 \\+{2\choose 1,0,1,0}r_1r_3+{2\choose 0,1,1,0} r_2r_3+{2\choose 0,0,1,1} r_3r_4+{2\choose 0,0,2,0}r_3^2\\+{2\choose 1,0,0,1}r_1r_4 +{2\choose 0,0,2,0}r_3^2$ making the simplification of multinomial coefficients we obtain $(r_1+r_2+r_3+r_4)^2 = r_1^2+r_2^2+r_3^2+r_4^2 +2(r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4 )\\ \therefore r_1^2+r_2^2+r_3^2+r_4^2= \left(\sum_{k=1}^4 r_k\right)^2-2\sum_{1\leq i<j\leq 4} r_ir_j =\left(-\frac{a_3}{a_4}\right)^2-2\left( \frac{a_2}{a_4}\right)= 7^2-4=45$ where $a_4=1,a_3=7,a_2=2,a_1=0,a_0=-1720$ are the coefficients of the given quartic equation.

The next easy way to solve is using Newton-Girard formula