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Answer:
[tex]x^3y^6 + 27z^{42} = (xy^2 + 3z^{14})(x^2y^4 - 3xy^2z^{14} + 9z^{28})[/tex]
Step-by-step explanation:
Given
[tex]x^3y^6 + 27z^{42[/tex]
Required
Factor
Express both terms as a cube
[tex]x^3y^6 + 27z^{42} = (xy^2)^3 + (3z^{14})^3[/tex]
Let
[tex]a =(xy^2)[/tex]
[tex]b = (3z^{14})[/tex]
So:
[tex]x^3y^6 + 27z^{42} = a^3 + b^3[/tex]
Using sum of cubes:
[tex]a^3 + b^3 = (a + b)(a^2 - ab + b^2)[/tex]
So:
[tex]x^3y^6 + 27z^{42} = (a + b)(a^2 - ab + b^2)[/tex]
Substitute in, values of a and b
[tex]x^3y^6 + 27z^{42} = (xy^2 + 3z^{14})((xy^2)^2 - (xy^2)*(3z^{14}) + (3z^{14})^2)[/tex]
[tex]x^3y^6 + 27z^{42} = (xy^2 + 3z^{14})(x^2y^4 - 3xy^2z^{14} + 9z^{28})[/tex]