Question

\(\large\tt \color{black}{a^3=abc+2,\\b^3=abc-3,\\c^3=abc+3,\\find~~the~~lowest~~number~~of~~a,~~b,~~c~~\\if~~a,~~b,~and~c~~are~~three~~real~~numbers}\)

Answer 1

\(\large\tt \color{black}{a. 3^{\frac{1}{3}}\\b. 9^{\frac{1}{3}}\\c. 2\\d.none~ of ~these}\)

Answer 2

the lowest number of a,b,c

Answer 3

does solving for abc and equating help us any?

Answer 4

but i need some easy method solving through options i mean

Answer 5

is the question clear?

Answer 6

a,b and c are three real numbers and i have to find the lowest of them

Answer 7

solving the equations only spins my head

Answer 8

Does it work if you solve everything for "abc", and then equate those equations together?

Answer 9

no i tried that but i cant found the right path,it only works when u substitute 3^(1/3) in b

Answer 10

have u find anything

Answer 11

I've been going pretty much round in circles with these equations. I did find one equation that seemed useful, but how to solve it I wasn't sure. Wolfram popped out \(b=3^{1/3}\) and \(b=-\dfrac{3^{2/3}}{2^{1/3}}\).

Answer 12

which was the solution to \(3b^3+11=3b(b^3+5)^{1/3}(b^3+6)^{1/3}+2\) but not sure how to solve that

Answer 13

ok'
if i have to put option in variable a,b and c and to save time'
how do i decide here in which of variable a,b or c
should i substitute options for trial and error

Answer 14

@aum

Answer 15

@cj49

Answer 16

If a, b and c are all positive, then b must be the smallest number because b^3 = abc - 3 (largest number being subtracted from abc). The next higher number would be a and the highest number would be c. Equate each answer choice to b and see if that works.

Answer 17

yes b works for option \(\large\tt \color{black}{a. 3^{\frac{1}{3}}}\)

Answer 18

Does it work for the other two options?

Answer 19

no it doesnt work

Answer 20

thanks very much for reviewing this

Answer 21

You are welcome.

Answer 22

Wait a minute.... Is the 1/3 in the answer choices exponents?

Answer 23

yes of cos

Answer 24

Then the lowest of \(\Large \sqrt[3]{3}, ~~\sqrt[3]{9}, ~~ 2\) is \(\Large \sqrt[3]{3}\) and that is the answer because with b set to \(\Large \sqrt[3]{3}\) you can solve for a and c and get real values which will be higher than b making b the lowest number.

Answer 25

Initially the answer choices looked like 3 and 1/3; 9 and 1/3 rather than exponents.

Answer 26

yes it gives both values of \(\large\tt \color{black}{a~~and~~c}\)

Answer 27

alright.

Answer 28

a3 = (abc + 2), b3 = (abc - 3), c3 = (abc + 3).

Multiply all three equations and keep abc = x.

x3 = (x + 2)(x - 3)(x + 3)
--> 2x2 - 9x - 18 = 0
--> (2x + 3)(x - 6) = 0
--> x = -3/2 or x = 6.
Taking abc = 6
and
noting that b is the smallest number
b3 = 6 - 3 = 3 --> b = 3^(1/3)