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06 Jun 2017, 02:10
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Knowing that x + y + z = 4
x^2 + x^2 + z^2 = 4 and
x^3 + y^3 + z^3 = 4
Solving the equation with complex numbers
I know that: 4 = x^2 + y^2 + z^2 = (x+y+z)^2 - (x*y + x*z + y*z). So (x*y + x*z + y*z) = 12
4 = x^3 + y^3 + z^3 = (x+y+z^3 - 3*(x^2*y + x^2*z + x*y^2 + x*z^2 + y^2*z + y*z^2) - 6*x*y*z.
Trying to find (x^2*y + x^2*z + x*y^2 + x*z^2 + y^2*z + y*z^2) =
x*(x*y + x*z) + y*(x*y + y*z) + z*(x*z + y*z). Knowing that x+y+z = 4, I'm using that info so:
x*(x*y + x*z + y*z) + y*(x*y + y*z + x*z) + z*(x*z + y*z + x*y) - x*y*z - x*y*z - x*y*z.
we get, (x*y + x*z + y*z)*(x+y+z) - 3*x*y*z
Once I find what x*y*z I'm getting stuck in, how do I find the x,y,z (that are allegedly complex numbers) ?
x^2 + x^2 + z^2 = 4 and
x^3 + y^3 + z^3 = 4
Solving the equation with complex numbers
I know that: 4 = x^2 + y^2 + z^2 = (x+y+z)^2 - (x*y + x*z + y*z). So (x*y + x*z + y*z) = 12
4 = x^3 + y^3 + z^3 = (x+y+z^3 - 3*(x^2*y + x^2*z + x*y^2 + x*z^2 + y^2*z + y*z^2) - 6*x*y*z.
Trying to find (x^2*y + x^2*z + x*y^2 + x*z^2 + y^2*z + y*z^2) =
x*(x*y + x*z) + y*(x*y + y*z) + z*(x*z + y*z). Knowing that x+y+z = 4, I'm using that info so:
x*(x*y + x*z + y*z) + y*(x*y + y*z + x*z) + z*(x*z + y*z + x*y) - x*y*z - x*y*z - x*y*z.
we get, (x*y + x*z + y*z)*(x+y+z) - 3*x*y*z
Once I find what x*y*z I'm getting stuck in, how do I find the x,y,z (that are allegedly complex numbers) ?
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06 Jun 2017, 10:07
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mihaitudor
Dear mihaitudor,
I'm happy to respond.
First of all, I assume you know that this math is several levels beyond anything on the GMAT. There are no complex numbers on the GMAT. This is 100% off topic for the GMAT Quant. I want that to be crystal clear to everyone who reads this post.
When I first looked at this problem, I played around with algebra, and saw that this quickly led to a morass. I was intrigued by the symmetry, so I started playing around with numbers. Obviously, the numbers couldn't be three symmetrical complex numbers, each 120 degrees apart, because those would sum to zero. After a little playing around with numbers, I found the solution.
x = 2
y = 1 + i
z = 1 - i
x + y + z = 4
\(x^2\) = 4
\(y^2\) = 2i
\(z^2\) = -2i
\(x^2 + y^2 + z^2 = 4\)
\(x^3\) = 8
\(y^3\) = -2 +2i
\(z^3\) = -2 - 2i
\(x^3 + y^3 + z^3 = 4\)
It makes sense that the solution are three relatively simple complex numbers. The deviant who created this problem no doubt noticed what happened with powers of (1 + i) and (1 - i) and figured he would make this stumper of a problem.
Does all this make sense?
Mike
06 Jun 2017, 22:40
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Note that this problem is too hard for GMAT. Just for anyone who enjoys solving Math.
We have
\(x+y+z=4 \quad \quad\) (1)
\(x^2+y^2+z^2=4 \quad\) (2)
\(x^3+y^3+z^3 = 4 \quad\) (3)
From (1) we have \((x+y+z)^2=16 \implies (x^2+y^2+z^2)+2(xy+yz+zx)=16 \implies xy+yz+zx=6 \quad\) (4)
Also, from (1) we have \(x+y=4-z \quad\) (5)
From (2) we have \(x^2+y^2=(x+y)^2 - 2xy=4-z^2 \)
From (5) we have \( (4-z)^2 - 2xy = 4-z^2 \implies xy=z^2-4z+6 \quad\) (6)
Now, from (3) we have
\(x^3+y^3=4-z^3 \implies (x+y)^3 - 3xy(x+y) = 4-z^3\)
from (5) and (6) we have
\((4-z)^3-3(z^2-4z+6)(4-z)=4-z^3\)
Shorten this expression, we have
\(z^3-4z^2+6z-4=0\)
We have
\(z^3-4z^2+6z-4 \\
=(z^3-4z^2+4z)+(2z-4) \\
=z(z-2)^2+2(z-2) \\
=(z-2)(z(z-2)+2) \\
=(z-2)((z-1)^2+1)\)
Hence \((z-2)((z-1)^2+1) = 0 \implies z=2\) or \((z-1)^2+1=0 \implies z=2, 1 \pm i\)
So the root of the problem is \((1+i, 1-i, 2)\)
We have
\(x+y+z=4 \quad \quad\) (1)
\(x^2+y^2+z^2=4 \quad\) (2)
\(x^3+y^3+z^3 = 4 \quad\) (3)
From (1) we have \((x+y+z)^2=16 \implies (x^2+y^2+z^2)+2(xy+yz+zx)=16 \implies xy+yz+zx=6 \quad\) (4)
Also, from (1) we have \(x+y=4-z \quad\) (5)
From (2) we have \(x^2+y^2=(x+y)^2 - 2xy=4-z^2 \)
From (5) we have \( (4-z)^2 - 2xy = 4-z^2 \implies xy=z^2-4z+6 \quad\) (6)
Now, from (3) we have
\(x^3+y^3=4-z^3 \implies (x+y)^3 - 3xy(x+y) = 4-z^3\)
from (5) and (6) we have
\((4-z)^3-3(z^2-4z+6)(4-z)=4-z^3\)
Shorten this expression, we have
\(z^3-4z^2+6z-4=0\)
We have
\(z^3-4z^2+6z-4 \\
=(z^3-4z^2+4z)+(2z-4) \\
=z(z-2)^2+2(z-2) \\
=(z-2)(z(z-2)+2) \\
=(z-2)((z-1)^2+1)\)
Hence \((z-2)((z-1)^2+1) = 0 \implies z=2\) or \((z-1)^2+1=0 \implies z=2, 1 \pm i\)
So the root of the problem is \((1+i, 1-i, 2)\)
