Question 524437
Note that {{{ 64 = 2^6 }}}, so
{{{ y^6 - 64z^6 = y^6 - (2z)^6 }}}
and
{{{ y^6 - (2z)^6 = ( y3 + (2z)^3 )*( y^3 - (2z)^3) }}}
{{{ ( y3 + (2z)^3 )*( y^3 - (2z)^3) = ( y^3 + 8z^3)*(y^3 - 8z^3) }}}
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This is the factoring, but you can't really "solve" this
because it's not an equation 
( you need an = sign in the original expression )