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.!