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

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

$\sqrt[3]{64}$ = y

y = 4.

:)