Question

Challenge problem for y'all.

Answer 1

Solve the system of equations:\[3x + y - 2z = 0\]\[4x - y - 3z = 0\]\[x^3 + y^3 + z^3 = 467\]

Answer 2

I've got a sweet, sweet solution to this.

Answer 3

(5,-1,7)

Answer 4

I guess we could also find some complex solutions but that'd get ..... a little complex :D

Answer 5

@nipunmalhotra93 no complex :)
3 deg equation + 1 deg both odd

Answer 6

It's trivial to do if you just solve it the standard slow way but I'm not seeing any obvious clever short-cut

Answer 7

Your "standard slow way" can be a pretty fast way to do the problem. What did you use?

Answer 8

^yeah it's quite trivial. @BSwan why not complex?

Answer 9

combining the first two equations you get \(7x=5z\). if you multiply the top by \(4\) and bottom by \(-3\) you get:$$12x+4y-8z=0\\-12x+3y+9z=0\\-7y=z$$so \(x=-5y\) as well. We can now substitute \(y=-\frac15x,z=\frac75x\) to get:$$x^3-\frac1{125}x^3+\frac{343}{125}x^3=467\\\frac{467}{125}x^3=467\\x^3=125\implies x=5$$ therefore \(y=-1,z=7\)

Answer 10

It's better to use the parametric equation of the line of intersection of the two planes. Takes less than a minute.

Answer 11

@nipunmalhotra93 take the cross product of their normals for the direction and then pick an arbitrary point satisfying both?

Answer 12

while that's obviously not hard I'm not sure it's 'better' or easier

Answer 13

the most obvious point satisfying both is \((0,0,0)\) and the direction is just the cross product of \((3,1,-2),(4,-1,-3)\)

Answer 14

Yes take the cross product. They both satisfy the origin. Then you get x, y and z as linear functions of the parameter t.

Answer 15

Yes, and then you can relate them all to \(t\) and solve like that. I don't see how that is effectively any different than what I did where \(t=x\). Besides, cross products are computationally more expensive for a human to do than to just easily do elimination in your head

Answer 16

@BSwan Here are two complex solutions:
(-5t,t,-7t), where t is
\[e ^{i \pi/3}, e ^{i 5\pi/3}\]

Answer 17

this does not apply for equation 3

Answer 18

if you want the complex roots just look here:$$x^3=125\\(x/5)^3=1$$so take the three roots of unity \(1,e^{2\pi i/3},e^{4\pi i/3}\) and you get for \(x\): $$x\in\{5,5e^{2\pi i/3},5e^{4\pi i/3}\}$$

Answer 19

in @nipunmalhotra93 he has \(t=-x\) hence the phase difference of \(\pi/3\) (as \(e^{i\pi}=-1\))

Answer 20

\[3x + y - 2z = 0\]\[4x - y - 3z =0\]\[\dfrac{x}{\begin{vmatrix}1& -2 \\ -1 & -3 \end{vmatrix}}=\dfrac{-y}{\begin{vmatrix} 3 & -2 \\ 4 & -3 \end{vmatrix}} = \dfrac{z}{\begin{vmatrix} 3&1\\ 4 & -1\end{vmatrix}}\]\[\Leftrightarrow \dfrac{x}{5}= \dfrac{y}{-1}=\dfrac{z}{7}=k\]\[\Leftrightarrow x = 5k, ~~y = -k, ~~ z = 7k \]\[(5k)^3 + (-k)^3 +(7k)^3 = 467 \Longleftrightarrow 467k^3 = 467 \Longleftrightarrow k=1\]

Answer 21

I wouldn't say this is better... but...

Answer 22

cross multiplication method = parametric form = oldrin's method above

all are equivalent and equally exciting xD

Answer 23

@oldrin.bataku it's not really something we can argue about.... I feel comfortable using parameters because... let's say that's a bit more neat :D

Answer 24

oops \(t=-\frac15x\) in @nipunmalhotra93 rather. then @ParthKohli has \(k=\frac15 x\)

Answer 25

I just think my method is cleaner than explicitly resorting to Cramer's rule or parameterizing the intersection of the planes since it's determinant- and cross-product-free :-p all doable in one's head

Answer 26

not that those determinants are hard to do in one's head but they're not the easiest to keep track of all at once

Answer 27

That's pretty subjective, but yes: a beginner would choose your method.

Answer 28

@BSwan eh?

Answer 29

well yes it's subjective, I did say 'I just think'. Personally I think it's a better method for both beginners and experienced... algebra II students (?) :-p this is an algebra II problem is it not? It's less convoluted and simpler

Answer 30

@oldrin.bataku it's just a matter of preference I guess... :)

Answer 31

@parth complex roots are fine aren't they? :O

Answer 32

@oldrin.bataku One day, I'm going to find a question that you wouldn't be able to do. One day. >:)

Answer 33

@ParthKohli @oldrin.bataku are you both familiar with abstract algebra?

Answer 34

I'm not.

Answer 35

haha there are plenty of problems I cannot do and yes I am @nipunmalhotra93

Answer 36

I see... just wanted to know that's it >:D