1. Factor $9x^2 - 144z^2$ completely. Step 1: Recognize this as a difference of squares: $a^2 - b^2 = (a-b)(a+b)$. Step 2: Write $9x^2$ as $(3x)^2$ and $144z^2$ as $(12z)^2$. Step 3: Apply the difference of squares formula: $$9x^2 - 144z^2 = (3x - 12z)(3x + 12z)$$ Step 4: Factor out common factors from each binomial: $$(3x - 12z) = 3(x - 4z), \, (3x + 12z) = 3(x + 4z)$$ Step 5: Combine the factors: $$9x^2 - 144z^2 = 3 \cdot 3 (x - 4z)(x + 4z) = 9(x - 4z)(x + 4z)$$ --- 2. Factor $1 + 8x^9$ completely. Step 1: Recognize this as a sum of cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. Step 2: Write $1$ as $1^3$ and $8x^9$ as $(2x^3)^3$. Step 3: Apply the sum of cubes formula: $$1 + 8x^9 = (1 + 2x^3)(1^2 - 1 \cdot 2x^3 + (2x^3)^2) = (1 + 2x^3)(1 - 2x^3 + 4x^6)$$ --- 3. Factor $x^4 - 10x^2 + 9$ completely. Step 1: Let $y = x^2$, then the expression becomes $y^2 - 10y + 9$. Step 2: Factor the quadratic: $$y^2 - 10y + 9 = (y - 9)(y - 1)$$ Step 3: Substitute back $y = x^2$: $$(x^2 - 9)(x^2 - 1)$$ Step 4: Factor each difference of squares: $$(x - 3)(x + 3)(x - 1)(x + 1)$$ --- 4. Factor $64n^6 - 16n^3 + 1$ completely. Step 1: Let $y = 4n^3$, then the expression becomes: $$y^2 - 4y + 1$$ Step 2: Factor the quadratic $y^2 - 4y + 1$ using the quadratic formula: $$y = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$$ Step 3: Write the factorization: $$(y - (2 + \sqrt{3}))(y - (2 - \sqrt{3}))$$ Step 4: Substitute back $y = 4n^3$: $$(4n^3 - 2 - \sqrt{3})(4n^3 - 2 + \sqrt{3})$$ --- 5. Factor $a^6 - 27r^3 - a^4 - 3a^2r - 9r^2$ completely. Step 1: Group terms: $$(a^6 - a^4) - (27r^3 + 3a^2r + 9r^2)$$ Step 2: Factor $a^4$ from the first group: $$a^4(a^2 - 1)$$ Step 3: Factor $3r$ from the second group: $$-3r(9r^2 + a^2 + 3r)$$ Step 4: Rearrange and try to factor as a quadratic in $a^2$ or $r$; this expression is complex and does not factor nicely with elementary methods. Step 5: Alternatively, rewrite as: $$a^6 - a^4 - 3a^2r - 9r^2 - 27r^3$$ Step 6: Recognize $a^6 - a^4 = a^4(a^2 - 1) = a^4(a - 1)(a + 1)$ but the rest is complicated. Step 7: Since no simple factorization is apparent, the expression is likely factored as is or requires advanced techniques beyond this scope. --- 6. Factor $x^4y^8 - 3x^2y^4 + 1$ completely. Step 1: Let $z = x^2y^4$, then the expression becomes: $$z^2 - 3z + 1$$ Step 2: Factor the quadratic $z^2 - 3z + 1$ using the quadratic formula: $$z = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$$ Step 3: Write the factorization: $$(z - \frac{3 + \sqrt{5}}{2})(z - \frac{3 - \sqrt{5}}{2})$$ Step 4: Substitute back $z = x^2y^4$: $$(x^2y^4 - \frac{3 + \sqrt{5}}{2})(x^2y^4 - \frac{3 - \sqrt{5}}{2})$$