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Thanks to Sean and Nick from New Zealand for the suggestion! I adapted this question from a 2020 test in New Zealand.

Solve for x, y, and z if:

x2yz = 2
y2xz = 3
z2xy = 5

In fact, solve the general case for unknown constants a, b, c:

x2yz = a
y2xz = b
z2xy = c

As usual, watch the video for a solution.

Beautiful Question

Or keep reading.
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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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Answer To Three Equations Quadratic Difference

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Order the equations as (i), (ii), (iii).

x2yz = a
y2xz = b
z2xy = c

We will eliminate the squared terms with a neat trick. Multiply (i) by y, (ii) by z, and (iii) by x.

x2yy2z = ay
y2zxz2 = bz
xz2x2y = cx

Sum the three equations. Miraculously all the terms on the left hand side cancel! So we get:

0 = ay + bz + cx
0 = (x, y, z) · (c, a, b)

But there’s another way we could have eliminated the square terms! Multiply (i) by z, (ii) by x, and (iii) by y.

x2zyz2 = az
xy2x2z = bx
yz2xy2 = cy

Again sum the three equations, and again all the terms on the left hand side magically cancel! So we get:

0 = az + bx + cy
0 = (x, y, z) · (b, c, a)

We have derived two equations showing that the vector (x, y, z) is orthogonal to both vectors (c, a, b) and (b, c, a). How can we find such a vector that is mutually orthogonal to two vectors? Easy, we use use the cross product! So let’s calculate the cross product of (c, a, b) and (b, c, a) to find a mutually orthogonal vector. Then we just need to find the magnitude of (x, y, z). So let’s calculate the cross product.

Now we multiply this by a scalar t to find the solution vector

(x, y, z)
= t(a2bc, b2ac, c2ab)
= (t(a2bc), t(b2ac), t(c2ab))

Let’s go back to the original equations.

x2yz = a
y2xz = b
z2xy = c

If a = b = c = 0, then substituting into the solution vector gives only one solution (x, y, z) = (0, 0, 0). (We don’t even have to solve for t).

So suppose not all three constants are equal to 0. For whichever constant is non-zero, we can use the corresponding equation to solve for t. Let’s suppose a is not zero, so we will use the first equation. We substitute the solution vector into the first equation

x2yz = a
(t(a2bc))2t(b2ac)t(c2ab)) = a
t2(a2bc)2t2(b2ac)(c2ab) = a
t2[(a2bc)2 – (b2ac)(c2ab)] = a

Now it is just a matter of simplifying and solving for t.

t2[a4 – 2a2bc + bc2bc2 + ab3 + ac2a2bc] = a
t2[a4 – 3a2bc + ab3 + ac3] = a

Since a ≠ 0 we can divide both sides by a to get:

t2(a3 + b3 + c3 – 3abc) = 1
t = ±1/√(a3 + b3 + c3 – 3abc)

Thus we have the solution set:

(x, y, z)
= ±(a2bc, b2ac, c2ab)/√(a3 + b3 + c3 – 3abc)

To see how it works, consider a specific system:

x2yz = 2
y2xz = 3
z2xy = 5

We have a = 2, b = 3, c = 5, so we can evaluate:

a2bc = -11
b2ac = -1
c2ab) = 19
√(a3 + b3 + c3 – 3abc) = √70

Thus we have the two solutions:

x = -11/√70
y = -1/√70
z = 19/√70

x = 11/√70
y = 1/√70
z = -19/√70

And that’s the answer!

References

Toppr
https://www.toppr.com/ask/en-ca/question/solve-x2-yz-a2-y2-zx/

Math StackExchange
https://math.stackexchange.com/questions/3960053/solve-three-variable-system-x2-yz-1-y2-xz-2-z2-xy-3

YouTube community post
https://www.youtube.com/post/Ugkxy-uD3uApCj0LwiMcZYznSNQj34VAXQ7v

Cross product idea
https://math.stackexchange.com/questions/2129989/how-to-get-ratio-of-a-b-c-from-2-equations-in-a-b-c

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