$a^{4} + a^{2} b^{2} + b^{4}$
If $(a + b) = 8, a b = 15$

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Solution

$a^{4} + a^{2} b^{2} + b^{4}$
Above terms can be written as,
$a^{4} + 2 a^{2} b^{2} + b^{4} - a^{2} b^{2}$
$(a^{2})^{2} + 2 a^{2} b^{2} + (b^{2})^{2} - (a b)^{2}$
$(a^{2} + b^{2})^{2} - (a b)^{2}$
$(a^{2} + b^{2} + a b) (a^{2} + b - a b)$
$a + b = 8, a b = 15$
So, $(a + b)^{2} = 8^{2}$
$a^{2} + 2 a b + b^{2} = 64$
$a^{2} + 2 (15) + b^{2} = 64$
$a^{2} + b^{2} + 30 = 64$
By transposing,
$a^{2} + b^{2} = 64 - 30$
$a^{2} + b^{2} = 34$
Then, $a^{4} + a^{2} b^{2} + b^{4}$
= $(a^{2} + b^{2} + a b) (a^{2} + b^{2} - a b)$
= $(34 + 15) (34 - 15)$
= $49 \times 19$
= $931$

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