Find $a+b+c$ to the nearest integer

Question

I am learning Solving Equations and encounter this problem. I have no clue how to proceed.

Suppose $a$, $b$, and $c$ are positive integers such that $ab$=18, $bc=24$, and $ac=48$. Find $a+b+c$ to the nearest integer.

Answer 1

To find $a$: $$\frac{ab\cdot ac}{bc}=a^2$$ $$\frac{18\cdot 48}{24}=a^2$$ $$a^2=36$$ $$a=6$$

To find $b$: $$\frac{ab\cdot bc}{ac}=b^2$$ $$\frac{18\cdot 24}{48}=b^2$$ $$b^2=9$$ $$b=3$$

To find $c$: $$\frac{ac\cdot bc}{ab}=c^2$$ $$\frac{48\cdot 24}{18}=c^2$$ $$c^2=64$$ $$c=8$$

Hence: $$a+b+c=6+3+8=17$$

Answer 2

Notice that $$abc(a+b+c)=a^2bc+ab^2c+abc^2=(ab)(ca)+(ab)(bc)+(ca)(bc)$$ Also $$(abc)^2=(ab)(bc)(ca)$$

$\therefore$ $$a+b+c\ =\ \frac{(ab)(ca)+(ab)(bc)+(ca)(bc)}{\sqrt{(ab)(bc)(ca)}}$$

Answer 3

Being lazy (I am), considering $$ab=18 \qquad bc=24 \qquad ac=48$$ proceed by successive eliminations that is to say $b=\frac{18}a$, $c=\frac{48}a$. Using the middle equation $$bc=\frac{18}a \times\frac{48}a=\frac{864}{a^2}=24\implies a^2=\frac{864}{24}=36\implies a=6$$ which implies $b=3$ and $c=8$.

Making the problem more general such as $$ab=X \qquad bc=Y \qquad ac=Z$$ and doing the same (assuming $X,Y,Z>0$), you should arrive to $$a=\frac{\sqrt{XZ}} {\sqrt{Y}}\qquad b=\frac{\sqrt{XY}} {\sqrt{Z}}\qquad c=\frac{\sqrt{YZ}} {\sqrt{X}}\implies a+b+c=\frac{XY+XZ+YZ}{\sqrt{XYZ}}$$ which will be solution for any values (integer or not) of $X,Y,Z$.