Factorise x^3/216-8y^3
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x^3/216-8y^3
= (x/6)^3 -(2y)^3
as you can see now it is in the form of a^3 - b^3 which equals (a-b)( a^2 + ab + b^2)
using that identity
{ (x/6) - 2y} { (x/6)^2 + (x/6)(2y) + (2y)^2 }
= { (x/6) - 2y} { (x^2/36) + (xy/3) + (4y^2) }
= (x/6)^3 -(2y)^3
as you can see now it is in the form of a^3 - b^3 which equals (a-b)( a^2 + ab + b^2)
using that identity
{ (x/6) - 2y} { (x/6)^2 + (x/6)(2y) + (2y)^2 }
= { (x/6) - 2y} { (x^2/36) + (xy/3) + (4y^2) }
Adway006:
Thank
I meant Great thanks
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Answered by
4
This is the answer for this question...
THE RULE USED IS:-
{x³-y³=(x-y) (x²+xy+y²) }
Hope it helps
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