Question

Find 3 numbers, as if the cube of the 1st number exceeds their product by 2, the cube of the second no. is smaller than their product by 3, and the cube of the third no. exceeds the product by 3.

Answer 1

Hi..
this question is all about breaking the qyestion in to parts and converting wirds in to equations :)

first..lets take the 3 numbers as "a","b","c"

now considering the 1st argument

a^3 - 2 = a×b×c ------(1)

considering the 2nd argument

b^3 +3 = a×b×c -----(2)

consid3ring the final argument

c^3 - 3 = a×b×c ------(3)

now all we have to do is use these 3 equation to find a,b,c

now..lets change the above 3 equation so that we can solve this :D

a^3 = abc +2 ------(A)
b^3 = abc - 3 ------(B)
c^3 = abc +3 ------(C)

now..heres something tricky

(A) × (B) ×(C)

a^3 × b^3 × c^3 = (abc +2)(abc -3)(abc +3)
(abc)^3 = (abc)^3 + 2(abc)^2 - 9abc - 18
2(abc)^2 - 9abc -18 =0
(2abc - 3)(abc + 6) =0

now....we have found the value of product of those 3 numbers :)

abc = 3/2 or abc = (-6)

when abc = 3/2

a^3 = (-1/2)
a = -1/ (2)^(1/3)

b^3 = 9/2
b = 3/(2)^(1/3)

c^3 = -3/2
c = - (3/2)^(1/3)

when abc = -6

a^3 = (-8)
a = -2

b^3 = -3
b = - 3^(1/3)

c^3 = -9
c = - 3

yep..thats alll..

hope this would help you