Expand: ${(y^{4} + {4 y}^{2} + 1)}^{2}$ using suitable identity:

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Solution

We know that ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
where $a = y^{4}, b = 4 y^{2}, c = 1$

${(y^{4} + 4 y^{2} + 1)}^{2}$

$= {(y^{4})}^{2} + {(4 y^{2})}^{2} + {(1)}^{2} + 2 \times y^{4} \times 4 y^{2} + 2 \times 4 y^{2} \times 1 + 2 \times 1 \times y^{4}$

$= y^{8} + 16 y^{4} + 1 + 8 y^{6} + 8 y^{2} + 2 y^{4}$

$= y^{8} + 8 y^{6} + 18 y^{4} + 8 y^{2} + 1$

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