$(a^{4} - 2 a^{2} b^{2} + 4)^{2}$

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Solution

$(a^{4} - 2 a^{2} b^{2} + 4)^{2}$
formula: $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
$∴ (a^{4} - 2 a^{2} b^{2} + 4)^{2} = (a^{4})^{2} + (- 2 a^{2} b^{2})^{2} + 4^{2} + 2 a^{4} (- 2 a^{2} b^{2}) + 2 a^{4} \cdot 4 + 2 \cdot 4 \cdot (- 2 a^{2} b^{2})$
$= a^{8} + 4 a^{4} b^{4} + 16 - 4 a^{6} b^{2} + 8 a^{4} - 16 a^{2} b^{2}$
$= a^{8} + 4 a^{4} b^{4} - 6 a^{6} b^{2} - 16 a^{2} b^{2} + 8 a^{4} + 16$

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a^{4} + 2 a^{2} b^{2} + b^{4}

a^{4} + 2 a^{2} b^{2} + b^{4}

a^{4} - 2 a^{2} b^{2} + b^{4}

a^{4} - 2 a^{2} b^{2} + b^{4}

Factorise

(i) a⁴ − b⁴

(ii) p⁴ − 81

(iii) x⁴ − (y + z)⁴

(iv) x⁴ − (x − z)⁴

(v) a⁴ − 2a²b² + b⁴

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