Question

factorise=8x cube-1 upon 27y cube

Answer 1

$\textbf{Answer :}$

Now,

$8 {x}^{3} - \frac{1}{ 27{y}^{3} } \\ \\ = {(2x)}^{3} - {( \frac{1}{3y}) }^{3} \\ \\ = (2x - \frac{1}{3y} )( {(2x)}^{2} + (2x \times \frac{1}{3y} ) + {( \frac{1}{3y}) }^{2} ) \\ \\ = (2x - \frac{1}{3y} )(4 {x}^{2} + \frac{2x}{3y} + \frac{1}{9 {y}^{2} } ),$

which is the required factorization.

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Answer 2

hey ,this is the answer for given questionfactorise=8x cube-1 upon 27y cube

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