# Let n(X*Y)=24, n(X*Z)=15, and n(Y*Z)=40. Find n(X*Y*Z).

Let $$\displaystyle{n}{\left({X}\cdot{Y}\right)}={24},{n}{\left({X}\cdot{Z}\right)}={15}$$, and $$\displaystyle{n}{\left({Y}\cdot{Z}\right)}={40}$$. Find $$n(X\cdot Y\cdot Z).$$

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Talisha

We know that: $$\displaystyle{n}{\left({X}\cdot{Y}\right)}={n}{\left({X}\right)}\cdot{n}{\left({Y}\right)}$$, $$\displaystyle{n}{\left({X}\cdot{Z}\right)}={n}{\left({X}\right)}\cdot{n}{\left({Z}\right)}$$, $$\displaystyle{n}{\left({Y}\cdot{Z}\right)}={n}{\left({Y}\right)}\cdot{n}{\left({Z}\right)}$$,
and
$$\displaystyle{x}{\left({X}\cdot{Y}\cdot{Z}\right)}={n}{\left({X}\right)}\cdot{n}{\left({Y}\right)}\cdot{n}{\left({Z}\right)}$$
Therefore, $$\displaystyle{n}{\left({X}\right)}\cdot{n}{\left({Y}\right)}={24},{n}{\left({X}\right)}\cdot{n}{\left({Z}\right)}={15}$$
and
$$\displaystyle{n}{\left({Y}\right)}\cdot{n}{\left({Z}\right)}={40}$$
Multiplying the three equations, we get: $$\displaystyle{\left({n}{\left({X}\right)}\right)}^{{2}}{\left({n}{\left({Y}\right)}\right)}^{{2}}{\left({n}{\left({Z}\right)}\right)}^{{2}}={24}\cdot{15}\cdot{40}={14400}$$
Thus, $$\displaystyle{\left({n}{\left({X}\right)}\cdot{n}{\left({Y}\right)}\cdot{n}{\left({Z}\right)}\right)}^{{2}}={14400}\to{n}{\left({X}\right)}\cdot{n}{\left({Y}\right)}\cdot{n}{\left({Z}\right)}={120}$$
Finally, using (*), $$\displaystyle{n}{\left({X}\cdot{Y}\cdot{Z}\right)}={120}$$