Question
Factorise:
8a3 - b3 - 4ax + 2bx
8a3 - b3 - 4ax + 2bx
✓
Answer
8a3 - b3 - 4ax + 2bx
= 8a3 - b3 - 2x(2a - b)
= (2a)3 - (b)3 - 2x(2a - b) Since a3 - b3 = (a - b)(a2 + a × b + b2)
= (2a - b)[(2a)2 + 2a × b + (b)2] - 2x(2a - b)
= (2a - b)(4a2 + 2ab + b2) - 2x(2a - b)
= (2a - b)(4a2 + 2ab + b2 - 2x)
= 8a3 - b3 - 2x(2a - b)
= (2a)3 - (b)3 - 2x(2a - b) Since a3 - b3 = (a - b)(a2 + a × b + b2)
= (2a - b)[(2a)2 + 2a × b + (b)2] - 2x(2a - b)
= (2a - b)(4a2 + 2ab + b2) - 2x(2a - b)
= (2a - b)(4a2 + 2ab + b2 - 2x)
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