Find three number of the first number exceeds that their product by 3 and the cube of third numbers exceed their product by 3.
Answers
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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
Answer:
Hi..
this question is all about breaking the question in to parts and converting words 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)
considring 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..here 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 all..
hope this would help you
Step-by-step explanation: