Another System of Equations?

First of all divide the first equation by the second one and you will see $a^2-ab+b^2=20$ . Subtracting this equation from $a^2+ab+b^2=45$ we obtain $2ab=25$ .

It is given that: $\begin{cases} a^4 + (ab)^2 + b^4 = 900 & ...(1) \\ a^2 + ab + b^2 = 45 & ...(2) \end{cases}$

From (1):

$\begin{aligned} a^4 + (ab)^2 + b^4 & = 900 \\ \Rightarrow (a^2+b^2)^2 - (ab)^2 & = 900 \\ (\color{#3D99F6}{a^2+b^2+ab})(a^2+b^2-ab) & = 900 \quad \quad \small \color{#3D99F6}{\text{We note that } a^2 + ab + b^2 = 45 \quad ...(2) } \\ \color{#3D99F6}{45}(a^2+b^2-ab) & = 900 \\ \Rightarrow a^2+b^2-ab & = 20 \quad ...(3) \\ & \\ (2)-(3): \quad 2ab & = \boxed{25} \end{aligned}$

An alternative solution:

${ a }^{ 4 }+{ ( ab ) }^{ 2 }+{ b }^{ 4 }=900\Rightarrow\frac{a^{6}-b^{6}}{a^{2}-b^{2}}=900\Rightarrow\frac{a^{3}-b^{3}}{a-b}\cdot\frac{a^{3}+b^{3}}{a+b}=900$

${ a }^{ 2 }+ab+{ b }^{ 2 }=45\Rightarrow\frac{a^{3}-b^{3}}{a-b}=45$

Dividing, we get $a^2-ab+b^2=20$ , subtracting, we obtain $2ab=25$ .

$(a^2 + ab + b^2)^2 = 45^2 \\\\ \underbrace{a^4 + a^2b^2 + b^4}_{900} + 2a^3b + 2a^2b^2 + 2ab^3 = 2025 \\\\ 2ab \cdot \underbrace{(a^2 + ab + b^2)}_{45} = 1125 \\\\ \boxed{2ab = 25}$