cube root of 54a^4 (simplify)?
回答 (5)
2009-08-04 5:16 pm
✔ 最佳答案
54 = 27x2 = 3³ x 2... right?and
a^4 = a³ x a
so...54 a^4 = 3³a³ x 2a
and the cube root of that = 3a x ³√(2a)
2009-08-04 5:14 pm
If a radical is not a perfect cube, break it down to a cube and a co-factor is possible. Here, it is possible.
= ³â54a^4
= (³â27) (³âa³) (³â2) (³âa) (Notice that ³â27 and ³âa³ are cubes)
= 3a ³â2a (Answer)
Hope this helps.
= ³â54a^4
= (³â27) (³âa³) (³â2) (³âa) (Notice that ³â27 and ³âa³ are cubes)
= 3a ³â2a (Answer)
Hope this helps.
2009-08-04 5:26 pm
³â(54a^4)
= ³â[(3^3 * a^3 * 2 * a)]
= ³â(3^3) * ³â(a^3) * ³â2 * ³âa
= 3 * a * ³â(2 * a)
= 3a³â(2a)
= ³â[(3^3 * a^3 * 2 * a)]
= ³â(3^3) * ³â(a^3) * ³â2 * ³âa
= 3 * a * ³â(2 * a)
= 3a³â(2a)
2009-08-04 5:17 pm
Factor the radicand, then take the cube roots of the factors.
â(54a⁴) = â(3³Ã2Ãa³Ãa)
          = â(3³) à â2 à â(a³) à â
           = 3a â(2a)
â(54a⁴) = â(3³Ã2Ãa³Ãa)
          = â(3³) à â2 à â(a³) à â
           = 3a â(2a)
2009-08-04 5:11 pm
3 2^(1/3) a^4^(1/3)
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