Question

Factor x3y6 + 27z42. (xy3 + 3z39)(x2y9 – 3xy3z39 + 9z1,521) (xy3 + 3z39)(x2y6 – 3xy3z39 + 9z78) (x3y6 + 27z42)(x6y12 – 27x3y6z42 + 729z84) (xy2 + 3z14)(x2y4 – 3xy2z14 + 9z28)

Answer 1

Answer:

$x^3y^6 + 27z^{42} = (xy^2 + 3z^{14})(x^2y^4 - 3xy^2z^{14} + 9z^{28})$

Step-by-step explanation:

Given

$x^3y^6 + 27z^{42$

Required

Factor

Express both terms as a cube

$x^3y^6 + 27z^{42} = (xy^2)^3 + (3z^{14})^3$

Let

$a =(xy^2)$

$b = (3z^{14})$

So:

$x^3y^6 + 27z^{42} = a^3 + b^3$

Using sum of cubes:

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

So:

$x^3y^6 + 27z^{42} = (a + b)(a^2 - ab + b^2)$

Substitute in, values of a and b

$x^3y^6 + 27z^{42} = (xy^2 + 3z^{14})((xy^2)^2 - (xy^2)*(3z^{14}) + (3z^{14})^2)$

$x^3y^6 + 27z^{42} = (xy^2 + 3z^{14})(x^2y^4 - 3xy^2z^{14} + 9z^{28})$

Answer 2

Answer:

its d

Step-by-step explanation: