Question 52755
The formula for factoring the difference of two perfect cubes is:
{{{a^3-b^3=(a-b)(a^2+ab+b^2)}}}
{{{y^3-8)=(y*y*y-2*2*2)}}}
Therefore we have:
{{{y^3-2^3}}}
a=y and b=2 Plug them into your formula and you have:
{{{y^3-8=(y-2)(y^2+y*2+2^2)}}}
{{{y^3-8=highlight((y-2)(y^2+2y+4))}}}
Happy Calculating!!!