Question

Factorize:
1.) 64x^3-y^3
2.) x^4+10x^3+35x^2+50x+24
3.) x^3/8-64-3x^2+24x

Answer 1

1) an obvious difference of cubes

64x^3 - y^3 = (4x-y)(16x^2 + 4xy + y^2)

2) x^4+10x^3+35x^2+50x+24
I tried ±1, ±2, ±3
Found x=-1, x=-2, and x=-3 to work
so the first three factors would be
(x+1)(x+2)(x+3)(......)
By the "common sense theorem) the last factor would have to be (x+4)

3. trying different things ....

Answer 2

3. x^3/8-64-3x^2+24x
= (1/8)(x^3 - 512 - 24x^2 + 192x)
= (1/8)(x^3 - 24x^2 + 192x - 512)

noticed that 8^3 = 512, became suspicious it might be perfect cube, so
tried (x-8)^3 and found it to be x^3 - 24x^2 + 192x -512

so
x^3/8-64-3x^2+24x
= (1/8)(x-8)^3