# How to solve?

#### Subhotosh Khan

##### Super Moderator
Staff member
Please follow the rules of posting in this forum, as enunciated at:

Hint:

Using the first (set) of equation/s, calculate: a*b*c = ?

#### Blala

##### New member
Please follow the rules of posting in this forum, as enunciated at:

Hint:

Using the first (set) of equation/s, calculate: a*b*c = ?
What is the compleate solution? I can't solve.

#### Subhotosh Khan

##### Super Moderator
Staff member
What is the compleate solution? I can't solve.
You are given:

a*b = b*c = c*a = 4

Now calculate:

(a*b) * (b*c) * (c*a) = ?

#### Jomo

##### Elite Member
What is the compleate solution? I can't solve.
We don't give complete solutions. Rather we help you get the solution.
Try to get a common denominator and show us your work
Do you know what the common denominator would be?

#### Jomo

##### Elite Member
This problem is very strange. I am embarrassed that not only can I not do this problem but I do not see any errors in my work.

Here goes: $$\displaystyle \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{4}{abc} + \dfrac{4}{abc} + \dfrac{4}{abc} = \dfrac{12}{abc} = \dfrac{1}{2} \\ so \\abc =24$$

Since $$\displaystyle ab = 4$$, we get $$\displaystyle c=6$$ But $$\displaystyle ac = 4$$ so $$\displaystyle b=6$$ (similarly we get $$\displaystyle a=6$$). But this contradicts $$\displaystyle ab =ac=bc = 4$$

2nd method. $$\displaystyle abc=24$$ Now $$\displaystyle \dfrac{ab}{bc}=\dfrac{a}{c} = 1$$ So $$\displaystyle a=c$$ Same argument that $$\displaystyle b=c$$ so $$\displaystyle a=b=c$$. This yields that $$\displaystyle a=b=c=2$$ (assuming positive numbers). Then we get $$\displaystyle \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} =\dfrac{1}{2}$$ which is also not true.

Now if a, b and c were all -2, then the sum would not be 1/2. If only one was -2 and the others both 2, than that would contradict that the product of any 2 would be 4.

What is going on here?

#### lookagain

##### Elite Member
This problem is very strange. I am embarrassed that not only can I not do this problem but I do not see any errors in my work.

Here goes: $$\displaystyle \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{4}{abc} + \dfrac{4}{abc} + \dfrac{4}{abc} = \dfrac{12}{abc} = \dfrac{1}{2} \\ so \\abc =24$$

Since $$\displaystyle ab = 4$$, we get $$\displaystyle c=6$$ But
It is ok with that.

Try to have a look at my solution

Hi! Common Denominator: (bc+ac+ab)/abc=1/2
Next abc=16 and a=4/b so c=4 then if a=4/c then a=1 and b=4. So the answer is 1+4+4=9

Math Science by Daniel Dallas

MATHScience Daniel Dallas, you're incorrect. greg1313 explained how you are wrong in the Math Help Forum about this.

This problem is faulty/impossible.

#### MATHScience Daniel Dallas

##### New member
Common Denominator: (bc+ac+ab)/abc=1/2
Next abc=24 . However if You express abc through initial conditions like a=4/b, b=4/c and c=4/a and multiply abc you will have that abc=64/(abc) from where abc=+-8, wich is a contradiction to original statement about abc=24.

Math Science by Daniel Dallas

#### Subhotosh Khan

##### Super Moderator
Staff member
This problem is very strange. I am embarrassed that not only can I not do this problem but I do not see any errors in my work.

Here goes: $$\displaystyle \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{4}{abc} + \dfrac{4}{abc} + \dfrac{4}{abc} = \dfrac{12}{abc} = \dfrac{1}{2} \\ so \\abc =24$$

Since $$\displaystyle ab = 4$$, we get $$\displaystyle c=6$$ But $$\displaystyle ac = 4$$ so $$\displaystyle b=6$$ (similarly we get $$\displaystyle a=6$$). But this contradicts $$\displaystyle ab =ac=bc = 4$$

2nd method. $$\displaystyle abc=24$$ Now $$\displaystyle \dfrac{ab}{bc}=\dfrac{a}{c} = 1$$ So $$\displaystyle a=c$$ Same argument that $$\displaystyle b=c$$ so $$\displaystyle a=b=c$$. This yields that $$\displaystyle a=b=c=2$$ (assuming positive numbers). Then we get $$\displaystyle \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} =\dfrac{1}{2}$$ which is also not true.

Now if a, b and c were all -2, then the sum would not be 1/2. If only one was -2 and the others both 2, than that would contradict that the product of any 2 would be 4.

What is going on here?
Strange is correct

abc = $$\displaystyle \sqrt{64}$$ = 8

Here goes: $$\displaystyle \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$$

$$\displaystyle = \dfrac{bc + ab + ca}{abc} = \dfrac{3*4}{abc} = \dfrac{12}{8} \ \ne {\frac{1}{2}}$$......................... The given value

So the problem cannot be solved as posted.

#### Jomo

##### Elite Member
Strange is correct

abc = $$\displaystyle \sqrt{64}$$ = 8

Here goes: $$\displaystyle \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$$

$$\displaystyle = \dfrac{bc + ab + ca}{abc} = \dfrac{3*4}{abc} = \dfrac{12}{8} \ \ne {\frac{1}{2}}$$......................... The given value

So the problem cannot be solved as posted.
How are you getting abc=8?

#### Subhotosh Khan

##### Super Moderator
Staff member
How are you getting abc=8?
a*b = b*c = a*c = 4

(a*b) * (b*c) * (a*c) = 4 * 4 * 4 = 64

(a * b * c)2 = 82

continue.....