8x3+y3+64-24xy factorize..?

`8x^3+y^3+64-24xy`

There are quite a number of solutions to this. If this is part of a question, you need to see which factorisation suits the question the best

You can have

`(8x^3+64)+(y^3-24xy)`
=`8(x^3+8)+y(y^2-24x)`
=`8(x+2)(x^2-2x+4)+y(y^2-24x)`

OR

`(8x^3-24xy)+(y^3+64)`
=`8x(x^2-3y)+(y+4)(y^2-4y+16)`

OR

`(8x^3+y^3)+(64-24xy)`
=`(2x+y)(4x^2-2xy+y^2)+8(8-3xy)`

If one is familiar with the following factorisation, then the

solution can easily be obtained :

`a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).`

`:. "The Expression (exp.)="8x^3+y^3+64-24xy`,

`=(2x)^3+(y)^3+(4)^3-3(2x)(y)(4)`,

`=(2x+y+4){(2x)^2+(y)^2+(4)^2-(2x)(y)-(y)(4)-(4)(2x)}`,

`=(2x+y+4)(4x^2+y^2+16-2xy-4y-8x)`, is the

desired factorisation!

Otherwise :

Suppose that, `2x+y=u`.

`:. (2x+y)^3=u^3`.

`:. (2x)^3+y^3+3(2x)(y)(2x+y)=u^3," &, as, "2x+y=u,`

`8x^3+y^3+6uxy=u^3,`

`or, 8x^3+y^3=u^3-6uxy`.

Adding `(64-24xy)` on both sides, we get,

`8x^3+y^3+64-24xy=u^3-6uxy+64-24xy,`

`={(4)^3+(u)^3}-6uxy-24xy`,

`={color(red)((4+u))(4^2-4u+u^2)}-6xycolor(red)((u+4)),......[because, m^3+n^3=(m+n)^3-3mn(m+n)]`,

`=(u+4){(16-4u+u^2)-6xy}`,

`={(2x+y)+4}{16-4(2x+y)+(2x+y)^2-6xy}......[because, u=(2x+y)]`,

`=(2x+y+4){16-8x-4y+4x^2+4xy+y^2-6xy}`,

`=(2x+y+4)(4x^2+y^2+16-8x-4y-2xy)`, as before!

Enjoy Maths.!