Is this problem hard or easy?

$\big(a^2+b^2\big)^3=\big(a^3-3ab^2\big)^2+\big(b^3-3a^2b\big)^2=1818^2+2222^2$ So $a^2+b^2=\sqrt[3]{8242408}=\boxed{202}$

Slightly motivated but similar solution as @Culver Kwan 's solution:

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Note that $(a-ib)^3=a^3-3ab^2+i(b^3-3ab^2)$ and similarly $(a+ib)^3=a^3-3ab^2-i(b^3-3ab^2)$

Now multiplying both of the above equation yields $(a^2+b^2)^3=(a^3-3ab^2)^2+(b^3-3a^2b)^2$ .

Hence the answer is $(1818^2+2222^2)^{1/3}=\boxed{202}$ .

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Note: $i=\sqrt{-1}$ denotes the imaginary unit.