m^2 + 7m + 49 = 0 , m^3 = ?

for starters, we can have the following quick and clean way

m^2 + 7m + 49 = 0

m^3 + 7m^2 + 49m = 0

and since

m^2 + 7m + 49 = 0

m^2 = -7m - 49

then we can have

m^3 + 7m^2 + 49m = 0

m^3 + 7(-7m - 49) + 49m = 0

m^3 - 49m - 343 + 49m = 0

m^3 = 343

at this point we can naively just say m = 7 since 7^3 = 343

but what will happen if we do so ?

m^2 + 7m + 49 = (7)^2 + 7(7) + 49 = 49 + 49 + 49 =/= 0

what is going on ?!?! QAQ

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let's allow ourselves to dive into the complex plane,

here we can have the following

assume that the original quadratic equation we have is

m^2 + km + k^2 = 0

then

m

= (-k +or- √( k^2 - 4(1)(k^2) ))/2

= (-k +or- √( -3k^2 ))/2

= (-k +or- ki√(3) )/2

= k(-1 +or- i√(3) )/2

= (k/2)(-1 +or- i√(3))

then

m^3 = (k/2)^3 × (-1 +or- i√(3))^3

and by using the binomial distribution theorem, we can have

m^3

= (k/2)^3 ×

[

1 × (-1)^3 × (+or- i√(3))^0 +

3 × (-1)^2 × (+or- i√(3))^1 +

3 × (-1)^1 × (+or- i√(3))^2 +

1 × (-1)^0 × (+or- i√(3))^3

]

= (k/2)^3 ×

[

1 × (-1) × (1 ) +

3 × ( 1) × (+or- i√(3) ) +

3 × (-1) × (-3 ) +

1 × ( 1) × (-or+ 3i√(3))

]

= (k^3/8) × [ - 1 +or- 3i√(3) + 9 -or+ 3i√(3)]

= (k^3/8) × [8]

= k^3

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in this case k = 7, and from the proof above, we can see m^3 = k^3 = 7^3 = 343

in this case, it is just pure sheer luck.