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    I know this may seem like an easy question but I would need some help.

    Factorise completely 4y^9-81x^6
    It does not say what it is equal to, I assume it equal to 0.
    The nearest I have got is a weird looking (y^1.5 -x)(Y^1.5 +x)(2y-9x^2)(2y+9x^2)
    Any help would be appreciated.
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4y^9 - 81x^6



let p = y^{\frac{9}{2}}



let q = x^3



4y^9 - 81x^6 = 4p^2 - 81q^2 = (2p - 9q)(2p + 9q)



To complete the factorisation substitute p and q for y and x.

    Can't think of anything else apart from difference of two squares
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    (Original post by TurkishMathsHelp)
    I know this may seem like an easy question but I would need some help.

    Factorise completely 4y^9-81x^6
    It does not say what it is equal to, I assume it equal to 0.
    The nearest I have got is a weird looking (y^1.5 -x)(Y^1.5 +x)(2y-9x^2)(2y+9x^2)
    Any help would be appreciated.
    I wouldn't have thought fractional powers were allowed for the variables.

    You could treat it as the difference of two cubes.
 
 
 
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