Question

125y3 + 64z3     Factor the sum or difference of two cubes?

Answer 1

[5y]³+[4z]³
by using identity a³+b³=a³+b³+3a²b+3ab²
[5y]³+[4z]³+3×[5y]²×[4z]+3×[5y]×[4z]²
125y³+64z³+225y²z+240yz²

Answer 2

$125y^3 + 64z^3$
$= (5y)^3 + (4z)^3$
$Using\ identity:\ a^3+b^3=(a+b)(a^2+b^2-ab)$
We have: $(5y+4z)[(5y)^2+(4z)^2-(5y)(4z)]$
$=(5y+4z)(25y^2+16z^2-40yz)$