Question

Factorise 

1.  \(a^ 4 - b^ 4\) 
2.  \(p^ 4 - 81\) 
3.  \(x^ 4 - (y + z)^ 4\) 
4.  \(x^ 4 - (x - z)^ 4\) 
5.  \(a ^4 - 2a ^2b^ 2 + b ^4\)

Answer 1

(i) \(a^4 - b^4 = (a^2 )^2 - (b^ 2 )^ 2\) 
= \((a^ 2 - b^ 2 ) (a^ 2 + b ^2) \)) 
= \((a - b) (a + b) (a^ 2 + b ^2 )\)

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(ii) \(p^ 4 - 81 = (p ^2 ) ^2 - (9)^2 \)
= \((p^ 2 - 9) (p ^2 + 9)\) 
= \([(p)^ 2 - (3)^2 ] (p ^2 + 9)\) 
= \((p - 3) (p + 3) (p^ 2 + 9)\)

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(iii) \(x^ 4 - (y + z) ^4 = (x ^2 )^ 2 - [(y +z) ^2 ] ^2\) 
= \([x^ 2 - (y + z) ^2 ] [x ^2 + (y + z) ^2 ]\) 
= \([x - (y + z)][ x + (y + z)] [x^ 2 + (y + z) ^2 ]\) 
= \((x - y - z) (x + y + z) [x^ 2 + (y + z) ^2 ]\)

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(iv) \(x^ 4 - (x - z)^ 4 = (x ^2 )^ 2 - [(x - z) ^2 ] ^2\) 
= \([x^ 2 - (x - z) ^2 ] [x ^2 + (x - z) ^2 ]\) 
= \([x - (x - z)] [x + (x - z)] [x^ 2 + (x - z) ^2 ]\) 
= \(z(2x - z) [x^ 2 + x ^2 - 2xz + z^ 2 ]\) 
= \(z(2x - z) (2x^ 2 - 2xz + z^ 2 )\)

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(v) \(a^ 4 - 2a 2b^ 2 + b ^4 = (a ^2 )^ 2 - 2 (a ^2 ) (b ^2 ) + (b ^2 ) ^2\) 
= \((a^ 2 - b^ 2 )^ 2\) 
= \([(a - b) (a + b)]^2\) 
= \((a - b) ^2 (a + b) ^2\)

Answer 2

To factorize the given expressions, we can use the following identities:

1. \(a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)\) which further simplifies to \( (a^2 + b^2)(a + b)(a - b) \)

2. \(p^4 - 81 = (p^2 + 9)(p^2 - 9)\) which further simplifies to \( (p^2 + 9)(p + 3)(p - 3) \)

3. \(x^4 - (y + z)^4 = (x^2 - (y + z)^2)(x^2 + (y + z)^2)\) which further simplifies to \( (x - y - z)(x + y + z)(x^2 + y^2 + z^2 + 2yz) \)

4. \(x^4 - (x - z)^4 = (x^2 - (x - z)^2)(x^2 + (x - z)^2)\) which further simplifies to \( z(2x - z)(2x + z)(2x^2 + z^2) \)

5. \(a^4 - 2a^2b^2 + b^4 = (a^2 - b^2)^2\) which further simplifies to \( (a - b)^2(a + b)^2 \)