Question

How to find `m^2 + 4n^2 =?` from `m^3 - 12mn^2 = 40` `4n^3-3m^2n=10`

`m^3 - 12mn^2 = 40`
`4n^3-3m^2n=10`

`m^2 + 4n^2 =?`

Answer 1

`m^2+4n^2=10 root(3)(2)`

Explanation:

I assume `m` and `n` are real
`{(m^3-12mn^2=40), (4n^3-3m^2n=10):}`

Factor out what we can:
`{(m(m^2-12n^2)=40), (n(4n^2-3m^2)=10):}`

I can see that `n^2` appears 2 times with coefficients 4 and 12, so I could try to substitute `k=2n` or `k^2=4n^2`:
`{(m(m^2-3*4n^2)=40), (n(4n^2-3m^2)=10):}`

`{(m(m^2-3color(red)(k^2))=40), (n(color(red)(k^2)-3m^2)=10):}`

and double the second equation just for taste
`{(m(m^2-3k^2)=40), (color(red)(k)(k^2-3m^2)=20):}`

Distribute
`{(m^3-3mk^2=40), (k^3-3m^2k=20):}`

It looks very symmetrical now and closely related to cube of a sum. The problem is that adding or subtracting will never give perfect cube, besause signs wont match.

Maybe complex numbers can help?

Multiply second equation by `i`

`{(m^3-3mk^2=40), (ik^3-3im^2k=20i):}`

Adding gives
#{:(m^3,-3im^2k,-3mk^2,+ik^3,=40+20i), (m^3,+3m^2(-i)k,+3m(-1)k^2,+(i)k^3,=40+20i), (m^3,+3m^2(-i)k,+3m(-i)^2k^2,+(-i)^3k^3,=40+20i):}#

`(m-ik)^3=40+20i=20(2+i)`

Subtracting gives
#{:(m^3,+3im^2k,-3mk^2,-ik^3,=40-20i), (m^3,+3m^2(i)k,+3m(-1)k^2,+(-i)k^3,=40-20i), (m^3,+3m^2(i)k,+3m(i)^2k^2,+(i)^3k^3,=40-20i):}#

`(m+ik)^3=40-20i=20(2-i)`

Finally we have
`{((m-ik)^3=20(2+i)), ((m+ik)^3=20(2-i)):}`

Multiply them together using difference of squares
`((m-ik)(m+ik))^3=400(2+i)(2-i)`
`(m^2+k^2)^3=400(4+1)`, where `k=2n`
`(m^2+4n^2)^3=2000=2*10^3`

Take cube root for the answer
`m^2+4n^2=10 root(3)(2)`

Answer 2

`m^2+4n^2=10 root(3)(2)`

Explanation:

It is a solution without using complex numbers. This solution is shorter and more direct, but it's less clear why it works.

Same as before until this point:
#{(m^3-3mk^2=40), (k^3-3m^2k=20):}#

Square both equations
#{((m^3)^2-2*m^3*3mk^2+(3mk^2)^2=1600), ((k^3)^2-2*k^3*3m^2k+(3m^2k)^2=400):}#

Simplify
#{(m^6-6m^4k^2+9m^2k^4=1600), (k^6-6m^2k^4+9m^4k^2=400):}#

Add them together
`m^6+3m^4k^2+3m^2k^4+k^6=2000`

Use cube of sum
`(m^2+k^2)^3=2000`, where `k=2n`
`(m^2+4n^2)^3=2000=2*10^3`

Take cube root for the answer
`m^2+4n^2=10 root(3)(2)`