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

1 Answer
May 13, 2015

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.

:)