Question

the HCF of (x4 - y4) and (x6 - y6) is? ​

Answer 1

Answer:

Both numbers have 1,2,3, and 6 as common factors but among these common factors,6 is the highest common factor that can divide both 12 and 18 without remainders. This means 6 is the hcf

Answer 2

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$\sf \: = \: let \: f(x) = ( {x}^{4} - y ^{4} ) \: \: \: \: \: \: \: \\ \sf \: = \: ( {x}^{2} - {y}^{2} )( {x}^{2} + {y}^{2} ) \: \: \: \: \: \: \: \: \: \: \\ \sf \: = \: (x - y)(x + y)( {x}^{2} + {y}^{2} ) \\ \sf \: and \: g(x) = ( {x}^{6} - {y}^{6} ) \: \: \: \: \: \: \: \: \:$

$\sf \: = ({x}^{3})^{2} - ({y}^{3})^{2} \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: = \: ( {x}^{3} + {y}^{3} )( {x}^{3} - {y}^{3} )$

$\sf \: = (x + y) ({x}^{2} - xy + {y}^{2} )(x - y)({x}^{2} - xy + {y}^{2} ) \\ \sf \: = (x - y)(x + y)( {x}^{2} - xy + {y}^{2} ) (x² + xy + y²)$

$\sf \: \: HCF \: of \: [f(x), g(x)] = (x - y) (x + y) \\ \sf \: ans = {x}^{2} - {y}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

hope it was helpful to you