Asked by Algebra 2
I have a question...
I have no idea how to turn this into SRform
2187^(1/3) - 2(24)^(1/3)
I saw I could do this
2187^(1/3) - 2(4*6)^(1/3)
2187^(1/3) - 4(6)^(1/3)
then I do this
2187^(1/3)/ (6)^(1/3)
and I don't get a whole number so I can conclude that...
I can never get a whole number(6)^(1/3) = 2187^(1/3) which I need to do becasue 2(24)^(1/3) SRform is 4(6)^(1/3) and I can not get a whole number(6)^(1/3) for the SRform of 2187^(1/3) so what do I do?
I really don't understand the point of SRform either
please show me how to do this problem step by step and the reasoning
thanks
I have no idea how to turn this into SRform
2187^(1/3) - 2(24)^(1/3)
I saw I could do this
2187^(1/3) - 2(4*6)^(1/3)
2187^(1/3) - 4(6)^(1/3)
then I do this
2187^(1/3)/ (6)^(1/3)
and I don't get a whole number so I can conclude that...
I can never get a whole number(6)^(1/3) = 2187^(1/3) which I need to do becasue 2(24)^(1/3) SRform is 4(6)^(1/3) and I can not get a whole number(6)^(1/3) for the SRform of 2187^(1/3) so what do I do?
I really don't understand the point of SRform either
please show me how to do this problem step by step and the reasoning
thanks
Answers
Answered by
bobpursley
2187^(1/3) - 2(24)^(1/3)
I notice 24 is a perfect cube (8) times 3.
2187^(1/3) - 2(8*3)^(1/3) so I look at the 2187 to see if an 3 will divide out. Wallah! 2187=729*3 = 243*3*3=81*3*3*3 that is enough, knowing 81 ia 3*27
Now it is reduced to ...
(3*3^3*3^3)^1/3-6(3)^1/3
or
(9-6)(3)^1/3= 3*3^1/3
I notice 24 is a perfect cube (8) times 3.
2187^(1/3) - 2(8*3)^(1/3) so I look at the 2187 to see if an 3 will divide out. Wallah! 2187=729*3 = 243*3*3=81*3*3*3 that is enough, knowing 81 ia 3*27
Now it is reduced to ...
(3*3^3*3^3)^1/3-6(3)^1/3
or
(9-6)(3)^1/3= 3*3^1/3
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