What is the cubed-root of (27a^9b^12c^6)? Please, help me.
2007-08-15 07:48:46 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(27 a^9 b^12 c^6) ^ (1/3)
3 a ³ b^4 c ²
2007-08-15 08:07:25 · answer #1 · answered by Como 7 · 2⤊ 0⤋
You can rewrite it as:
(27a^9b^12c^6)^(1/3)
Then you can simply multiply the exponents by (1/3) AND taking the third root of each coefficient:
3a^3b^4c^2
2007-08-15 07:56:36 · answer #2 · answered by miggitymaggz 5 · 0⤊ 0⤋
cubed-root of (27a^9b^12c^6)
= (27a^9b^12c^6)^(1/3)
= 3 a^3 b^4 c^2
2007-08-15 07:53:21 · answer #3 · answered by vlee1225 6 · 0⤊ 0⤋
3(a^3)(b^4)(c^2)
You just take the cube root of each factor. The cube root of x^(3a) is x^a. So, for example, the cube root of b^12 is b^4.
2007-08-15 07:53:06 · answer #4 · answered by DavidK93 7 · 0⤊ 0⤋
3a^3b^4c^2
3*3*3=27
9/3=3
12/3=4
6/3=2
Notice that the exponents are divided by 3 in a cubed-root.
2007-08-15 07:54:46 · answer #5 · answered by Scruff 1 · 0⤊ 0⤋