Identities!

The interesting thing to note here is $(a^2 + b^2)^3 = (a^3 - 3ab^2)^2 + (b^3 -3ba^2)^2$ . But how does one come about the result? The motivation to link the familiar looking terms from the expansion of $(a \pm b)^3$ to what we need to find using complex numbers.

$\displaystyle \begin{array}{c}\\ (a^2 + b^2)^3 && = \left( (a + ib)(a-ib) \right)^3 \\ && = ( a + ib )^3 (a - ib )^3 \\ && = \left( (a^3 - 3ab^2) - i(b^3 - 3ba^2) \right) \left( (a^3 - 3ab^2) + i(b^3 - 3ba^2) \right) \\ && = (a^3 - 3ab^2)^2 + (b^3 - 3ba^2)^2 \\ \end{array}$

The last step uses the fact that $z \bar{z} = |z|^2$ for any complex number $z$ .

Plug in the given values in the above relation to obtain the required value as $\boxed{5}$ .

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

From the problem, we can get:

${ (a }^{ 3 }-3a{ b }^{ 2 })^{ 2 }+(b^{ 3 }-3{ a }^{ 2 }b)^{2}={ 5 }^{ 2 }+{ 1 }0^{ 2 }=125$

Expanding by algebraic identities, we have:

${ a }^{ 6 }-6{ a }^{ 4 }{ b }^{ 2 }+9{ a }^{ 2 }{ b }^{ 4 }+{ b }^{ 6 }-6{ a }^{ 2 }{ b }^{ 4 }+9{ a }^{ 4 }{ b }^{ 2 }=125$

Simplifying yields:

${ a }^{ 6 }+{ b }^{ 6 }+3{ a }^{ 4 }{ b }^{ 2 }+3{ a }^{ 2 }{ b }^{ 4 }=125={ 5 }^{ 3 }$

Factoring by algebraic identities, we gain:

$({ a }^{ 2 }+{ b }^{ 2 })^{ 3 }={ 5 }^{ 3 }$

Because $a$ and $b$ are real numbers, we can easily conclude:

$({ a }^{ 2 }+{ b }^{ 2 })={ 5 }$