Factor: 1. \( m^{3}+27 \) [A] \( (m+3)^{3} \) [B] \( (m+3)\left(m^{2}+9\right) \) \( [C](m+3)\left(m^{2}-3 m+9\right) \) [D] \( (m-3)\left(m^{2}+3 m+9\right) \) 2. \( 8 b^{3}-125 \) [A] \( (2 b-5)\left(4 b^{2}+10 b+25\right) \) [B] \( (2 b-5)^{3} \) \( [C](2 b-5)\left(4 b^{2}+25\right) \) [D] \( (2 b+10)\left(4 b^{2}-5 b+25\right) \) 3. \( 27 e^{3}-8 \) \( [\mathrm{A}](3 e+6)\left(9 e^{2}-2 e+4\right) \) [B] \( (3 e-2)^{3} \) [C] \( (3 e-2)\left(9 e^{2}+6 e+4\right) \) [D] \( (3 e-2)\left(9 e^{2}+4\right) \) 4. \( 8 r^{3}+125 \) \( [A](2 r+5)\left(4 r^{2}+25\right) \) [B] \( (2 r+5)\left(4 r^{2}-10 r+25\right) \) [C] \( (2 r+5)^{3} \) [D] \( (2 r-10)\left(4 r^{2}+5 r+25\right) \) \[ u^{3}-216 \] [A] \( (u-6)^{3} \) [B] \( (u-6)\left(u^{2}+36\right) \) \( [C](u-6)\left(u^{2}+6 u+36\right) \) D] \( (u+6)\left(u^{2}-6 u+36\right) \) 6. \( 27 d^{3}-64 \) [A] \( (3 d-4)\left(9 d^{2}+12 d+16\right) \) [B] \( (3 d+12)\left(9 d^{2}-4 d+16\right) \) [C] \( (3 d-4)^{3} \) [D] \( (3 d-4)\left(9 d^{2}+16\right) \) 7. \( c^{3}+8 \) [A] \( (c+2)\left(c^{2}+4\right) \) [B] \( (c-2)\left(c^{2}+2 c+4\right) \) \( [\mathrm{C}](c+2)\left(c^{2}-2 c+4\right) \) [D] \( (c+2)^{3} \) 8. \( 8 z^{3}+125 \) [A] \( (2 z+5)\left(4 z^{2}+25\right) \) [B] \( (2 z+5)^{3} \) [C] \( (2 z-10)\left(4 z^{2}+5 z+25\right) \) [D] \( (2 z+5)\left(4 z^{2}-10 z+25\right) \) 9. \( g^{3}-8 \) \( [\mathrm{A}](g-2)\left(g^{2}+4\right) \) [B] \( (g-2)^{3} \) [C] \( (g-2)\left(g^{2}+2 g+4\right) \) [D] \( (g+2)\left(g^{2}-2 g+4\right) \) 10. \( b^{3}+64 \) \( [A](b+4)\left(b^{2}-4 b+16\right) \) [B] \( (b+4)\left(b^{2}+16\right) \) [C] \( (b-4)\left(b^{2}+4 b+16\right) \) [D] \( (b+4)^{3} \)

1. We have \[ m^3 + 27 = m^3 + 3^3. \] Using the sum of cubes formula \[ a^3 + b^3 = (a+b)\left(a^2 - ab + b^2\right), \] set \(a=m\) and \(b=3\). Then it factors as \[ (m+3)\left(m^2-3m+9\right). \] Thus, the answer is [C]. 2. We have \[ 8b^3-125 = (2b)^3 - 5^3. \] Using the difference of cubes formula \[ a^3-b^3=(a-b)\left(a^2+ab+b^2\right), \] with \(a=2b\) and \(b=5\), we get \[ (2b-5)\left((2b)^2+2b\cdot5+5^2\right) = (2b-5)(4b^2+10b+25). \] Thus, the answer is [A]. 3. We have \[ 27e^3-8 = (3e)^3-2^3. \] Using the difference of cubes formula with \(a=3e\) and \(b=2\): \[ (3e-2)\left((3e)^2+3e\cdot2+2^2\right) = (3e-2)(9e^2+6e+4). \] Thus, the answer is [C]. 4. We have \[ 8r^3+125 = (2r)^3+5^3. \] Using the sum of cubes formula with \(a=2r\) and \(b=5\): \[ (2r+5)\left((2r)^2-2r\cdot5+5^2\right) = (2r+5)(4r^2-10r+25). \] Thus, the answer is [B]. 5. We have \[ u^3-216 = u^3-6^3. \] Using the difference of cubes formula with \(a=u\) and \(b=6\): \[ (u-6)\left(u^2+6u+36\right). \] Thus, the answer is [C]. 6. We have \[ 27d^3-64 = (3d)^3-4^3. \] Using the difference of cubes formula with \(a=3d\) and \(b=4\): \[ (3d-4)\left((3d)^2+3d\cdot4+4^2\right) = (3d-4)(9d^2+12d+16). \] Thus, the answer is [A]. 7. We have \[ c^3+8 = c^3+2^3. \] Using the sum of cubes formula with \(a=c\) and \(b=2\): \[ (c+2)\left(c^2-2c+4\right). \] Thus, the answer is [C]. 8. We have \[ 8z^3+125 = (2z)^3+5^3. \] Using the sum of cubes formula with \(a=2z\) and \(b=5\): \[ (2z+5)\left((2z)^2-2z\cdot5+5^2\right) = (2z+5)(4z^2-10z+25). \] Thus, the answer is [D]. 9. We have \[ g^3-8 = g^3-2^3. \] Using the difference of cubes formula with \(a=g\) and \(b=2\): \[ (g-2)\left(g^2+2g+4\right). \] Thus, the answer is [C]. 10. We have \[ b^3+64 = b^3+4^3. \] Using the sum of cubes formula with \(a=b\) and \(b=4\): \[ (b+4)\left(b^2-4b+16\right). \] Thus, the answer is [A].