what is the cube root of -128 in simplified form?
回答 (4)
2015-01-09 6:58 am
- 128 = 128 e^ (i(2n pi +pi))
Now take the cube root
(-128)^(1/3) = 4* 2^(1/3) ( e^(i/3)(2 npi +pi))
n=0
one root is 4 *2(1/3) ( cos pi/3 + i sin pi/3)= 2*2^(1/3) ( 1 + i root 3)
n=1 is the second root
= 4* 2^(1/3) (cos 180 + i sin 180)
n=3 is the third root
= 4 * 2^(1/3) ( cos 240 + i sin 240)
Now take the cube root
(-128)^(1/3) = 4* 2^(1/3) ( e^(i/3)(2 npi +pi))
n=0
one root is 4 *2(1/3) ( cos pi/3 + i sin pi/3)= 2*2^(1/3) ( 1 + i root 3)
n=1 is the second root
= 4* 2^(1/3) (cos 180 + i sin 180)
n=3 is the third root
= 4 * 2^(1/3) ( cos 240 + i sin 240)
2015-01-09 3:02 am
-128 = (-2)^7, then cube root (-128) = (-2)^2 * cube root (-2)
= 4 cube root (-2)
this can be converted into 4 cube root (-1 * 2)
= -4 cube root 2 ---- so either answer is correct
= 4 cube root (-2)
this can be converted into 4 cube root (-1 * 2)
= -4 cube root 2 ---- so either answer is correct
2015-01-09 2:51 am
-4 times the cube root of 2. we know that -128 is -2^7. if we factor out -2^6 from (-2^7)^(1/3) then we get -4(2^(1/3)).
2015-01-09 2:46 am
-128 = -2^7 so curt(-128) = -2^2 * curt(2) = -4curt(2)
where curt is cube root
where curt is cube root
收錄日期: 2021-04-21 01:11:18
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