Question

How do you factor `192^3 - 3y^3`?

Answer 1

Supposing 192³ - 3y³ equals 0,

3 x `2^6` - 3y³ = 0

Number 3 is multiplying both `2^6` and y³, so it is shorter to write

3(`2^6` - y³) = 0

Just for curiosity's sake, we can solve this for y: pass the 3 to the other side, dividing zero, which will obviously end up equaling zero. Then, we'd have left

`2^6` - y³ = 0

`2^6` = y³

64 = y³

`root(3)64` = y

y = 4.

:)