Question 75314
Factor"
{{{64a^3c^6+x^6y^9}}} Do you recognise this as the sum of two cubes?
{{{(4ac^2)^3+(x^2y^3)^3}}}
The sum of two cubes can be factored thus:
{{{A^3+B^3 = (A+B)(A^2-AB+B^2)}}} Applying this to your expression, where:{{{A = 4ac^2}}} and {{{B = x^2y^3}}}, we get:
{{{64a^3c^6+x^6y^9 = (4ac^2+x^2y^3)(16a^2c^4-4ac^2x^2y^3+x^4y^6)}}}