Solution:

Given,

$\text{1^{st} expression} = a^4 + a^2b^2 +b^4$

$\implies (a^2)^2 + (b^2)^2 + a^2b^2$

$\implies (a^2 + b^2)^2 - 2a^2b^2 + a^2b^2$

$\implies (a^2 + b^2)^2 - a^2b^2$

$\implies (a^2 + b^2 - (ab)^2$

[ Using $a^2 - b^2 = (a+b)(a-b)$ ]

$\implies (a^2 +ab +b^2)(a^2 - ab + b^2)$


$\text{2^{nd} expression} = a^3 + b^3$

$\implies (a +b)(a^2 -ab + b^2)$


$\text{3^{rd} expression} = a^3 - a^2b + ab^2$

$\implies a(a^2 - ab + b^2)$


We know,

$\text{Highest Common Factor (H.C.F.) = common factors only$

$\implies \text{H.C.F.} = (a^2 -ab+b^2)$


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