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Question
Factorise:
`16a^3 - 128/b^3`
Sum
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Solution
It is known that,
a3 − b3 = (a − b)(a2 + ab + b2)
`16a^3 - 128/b^3`
= `16[a^3 - 8/b^3]`
= `16[(a)^3 - (2/b)^3]`
= `16[(a - 2/b) {(a)^2 + (a) xx (2/b) + (2/b)^2}]`
= `16(a - 2/b) (a^2 + (2a)/b + 4/b^2)`
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