Relatively easy factoring algebra two problems I can't seem to get?

Question

Here they are:

factor

x^3-17x^2+72x

81x^6+72x^3y+16y^2

x^4y^4-9y^4

216x^3-1

125a^3-216

x^2-2xy-99y^2

8d^3+27

m^3-1000n^6

8n^3+229

please help?! most will be "special factoring" or whatever... thanks!

Answer 1

x^3-17x^2+72x=x(x^2-17x+72)=x(x-8)(x-9) =>common factor

81x^6+72x^3y+16y^2=(9x^3+4y)^2 =>perfect square

x^4y^4-9y^4=y^4(x^2+3)(x^2-3) =>common factor and difference of two squares

216x^3-1=(6x)^3-1=(6x-1)(36x^2+6x+1) =>factoring the cubic x^3-1=(x-1)(x^2+x+1)

125a^3-216=(5a)^3-6^3=(5a-6)(25a^2+30a+36) =>factoring the cubic x^3-y^3=(x-y)(x^2+xy+y^2)

x^2-2xy-99y^2=(x+9y)(x-11y) =>no trick

8d^3+27=(2d)^3+3^3=(2d+3)(4d^2-6d+9) =>factoring the cubic x^3+y^3=(x+y)(x^2-xy+y^2)

m^3-1000n^6=(m-10n^2)(m^2+10mn^2+100n^4) =>see x^3-y^3 above

8n^3+229 doesn't factor nicely (by writing y^3=229, so that y~6.12, we can factor this as 8n^3+229=8n^3+y^3=(2n+y)(4n^2-2ny+y^2)

Answer 2

one million:) a² - 5ab - 36b² =a² - 9ab + 4ab - 36b² [ using fact 36 = 6 x 6 = a(a- 9b) + 4b( a - 9b) =2 x3 x 2 x 3 = (a + 4b) ( a - 9b) answer = 4 x 9 =36 additionally 9 -4 =5 ] 2:) x² + 6xb + 9b² (x + 3b) ( x+ 3b) [ clean up direct using fact 3 x 3 =9 and 3+3=6] subsequently if the 1st term is x² in basic terms x² then you are able to clean up it immediately

Answer 3

x(x - 8) (x - 9)

Perfect square

Difference of two squares

Difference of two cubes

Difference of two cubes

Difference of two squares

Sum of two cubes

Difference of two cubes

Sum of the two cubes