Factor x^3y^6+27z^42

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

Rewrite $x^{3} y^{6}$ as ${(x y^{2})}^{3}$ .

${(x y^{2})}^{3} + 27 z^{42}$

Rewrite $27 z^{42}$ as ${(3 z^{14})}^{3}$ .

${(x y^{2})}^{3} + {(3 z^{14})}^{3}$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x y^{2}$ and $b = 3 z^{14}$ .

$(x y^{2} + 3 z^{14}) ({(x y^{2})}^{2} - (x y^{2}) (3 z^{14}) + {(3 z^{14})}^{2})$

Simplify.

Apply the product rule to $x y^{2}$ .

$(x y^{2} + 3 z^{14}) (x^{2} {(y^{2})}^{2} - (x y^{2}) (3 z^{14}) + {(3 z^{14})}^{2})$

Multiply the exponents in ${(y^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$(x y^{2} + 3 z^{14}) (x^{2} y^{2 \cdot 2} - (x y^{2}) (3 z^{14}) + {(3 z^{14})}^{2})$

Multiply $2$ by $2$ .

$(x y^{2} + 3 z^{14}) (x^{2} y^{4} - (x y^{2}) (3 z^{14}) + {(3 z^{14})}^{2})$

Remove parentheses.

$(x y^{2} + 3 z^{14}) (x^{2} y^{4} - x y^{2} (3 z^{14}) + {(3 z^{14})}^{2})$

Multiply $3$ by $- 1$ .

$(x y^{2} + 3 z^{14}) (x^{2} y^{4} - 3 x y^{2} z^{14} + {(3 z^{14})}^{2})$

Apply the product rule to $3 z^{14}$ .

$(x y^{2} + 3 z^{14}) (x^{2} y^{4} - 3 x y^{2} z^{14} + 3^{2} {(z^{14})}^{2})$

Raise $3$ to the power of $2$ .

$(x y^{2} + 3 z^{14}) (x^{2} y^{4} - 3 x y^{2} z^{14} + 9 {(z^{14})}^{2})$

Multiply the exponents in ${(z^{14})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$(x y^{2} + 3 z^{14}) (x^{2} y^{4} - 3 x y^{2} z^{14} + 9 z^{14 \cdot 2})$

Multiply $14$ by $2$ .

$(x y^{2} + 3 z^{14}) (x^{2} y^{4} - 3 x y^{2} z^{14} + 9 z^{28})$

Factor x^3y^6+27z^42

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