factor completely
2006-07-31 01:44:47 · 4 answers · asked by someone 2 in Science & Mathematics ➔ Mathematics
54x^6y^3-38x^3y^3 rather.....
2006-07-31 01:57:19 · update #1
54x^6y^3-38x^3y^3 = (54x^3 - 38)x^3y^3 = etc you can also factorout the gcd of (54,38)
2006-07-31 02:23:46 · answer #1 · answered by gjmb1960 7 · 1⤊ 0⤋
2x^3y^3(27x^3-19)
You could break down the second factor if you treated it like a perfect cube. It would be ugly and I don't recommend it.
Most instructors would have you stop here unless this was an exceptionally hard question and you are supposed to be studying how to factor things that look like: x^3-y^3.
2006-07-31 02:17:57 · answer #2 · answered by tbolling2 4 · 0⤊ 0⤋
54x^6y^3 - 38x^3y^3 = (2x^3y^3)(27x^3 - 19)
2006-07-31 03:45:38 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
2x3x3x17x37xy
2006-07-31 01:49:35 · answer #4 · answered by know it all 1 · 0⤊ 0⤋