I don't know how to solve this

[austral]

Hi everyone. So, I'm going to take an important maths exam next month (it could change my life forever) and I'm studying maths by myself with some past exam papers. I'm trying to understand every problem in every exam, and the topics vary a lot. I came across this one:

We are to find the value of x⁴+y⁴+z⁴ when x, y and z are real numbers which satisfy the following three equalities:

x+y+z = 3
x²+y²+z² = 9
xyz = -2

Firstly, it follows from the first two equalities that:

xy+yz+zx = A

Next, using

(x²+y²+z²)² = x⁴+y⁴+z⁴+B = {(xy)²+(yz)²+(zx)²}

we have

x⁴+y⁴+z⁴ = C

I need to find A, B and C.

I don't know where to start. It's the first time I find a problem like this in the papers. Also, I don't really understand the "it follows from the first two equalities that..." part.

And how are these types of problems called? I'm asking so I can search for more to practice for the exam.

Thanks in advance.

 

[Deleted member 4993]

Hi everyone. So, I'm going to take an important maths exam next month (it could change my life forever) and I'm studying maths by myself with some past exam papers. I'm trying to understand every problem in every exam, and the topics vary a lot. I came across this one:

We are to find the value of x⁴+y⁴+z⁴ when x, y and z are real numbers which satisfy the following three equalities:

x+y+z = 3
x²+y²+z² = 9
xyz = -2

Firstly, it follows from the first two equalities that:

xy+yz+zx = A

Next, using

(x²+y²+z²)² = x⁴+y⁴+z⁴+B = {(xy)²+(yz)²+(zx)²}

we have

x⁴+y⁴+z⁴ = C

I need to find A, B and C.

I don't know where to start. It's the first time I find a problem like this in the papers. Also, I don't really understand the "it follows from the first two equalities that..." part.

And how are these types of problems called? I'm asking so I can search for more to practice for the exam.

Thanks in advance.

(x + y + z)^2 = (x^2 + y^2 + z^2) + 2 * (xy + yz + zx)

Continue....

 

[Dr.Peterson]

Firstly, it follows from the first two equalities that:

xy+yz+zx = A

To find this, square the first equation you were given, expand, and compare the result to the second equation.

Next, using

(x²+y²+z²)² = x⁴+y⁴+z⁴+B = {(xy)²+(yz)²+(zx)²}

I think you typed that wrong, and it should be (x²+y²+z²)² = x⁴+y⁴+z⁴+B{(xy)²+(yz)²+(zx)²}.

To get this, expand the square on the left-hand side, as you will have done with the first equation.

Then you'll want to find what (xy)²+(yz)²+(zx)² is, for which you'll find it helpful to square xy+yz+zx.

we have

x⁴+y⁴+z⁴ = C

Plug in two of the equations you'll already have into that last one.

And how are these types of problems called? I'm asking so I can search for more to practice for the exam.

These may help as a start:

Symmetric Polynomials | Brilliant Math & Science Wiki

A symmetric polynomial is a polynomial where if you switch any pair of variables, it remains the same. For example, ...

brilliant.org

Art of Problem Solving

artofproblemsolving.com

It looks like your problem is trying to guide you through the process of working with such sums.

 

[austral]

x+y+z = 3
x²+y²+z² = 9
xyz = -2
xy+yz+zx = A

(x+y+z)² = x²+y²+z²+2xy+2yz+2xz
(x+y+z)²- (x²+y²+z²) .......................... edited - need those parentheses
(3)²-9
9-9
0

A=0

(x²+y²+z²)² = x⁴+y⁴+z⁴+B{(xy)²+(yz)²+(zx)²}

x⁴+y⁴+z⁴+2x²y²+2y²z²+2x²z² = x⁴+y⁴+z⁴+B{(xy)²+(yz)²+(xz)²}

x⁴+y⁴+z⁴+2x²y²+2y²z²+2x²z² = x⁴+y⁴+z⁴+B{x²y²+y²z²+x²z²}

x⁴+y⁴+z⁴+2x²y²+2y²z²+2x²z² = x⁴+y⁴+z⁴+2{x²y²+y²z²+x²z²}

x⁴+y⁴+z⁴+2x²y²+2y²z²+2x²z² = x⁴+y⁴+z⁴+2{xy+yz+xz)²-2xyz(x+y+z)

x⁴+y⁴+z⁴+2x²y²+2y²z²+2x²z² = x⁴+y⁴+z⁴+2{-2(-2)(3)}

(x²+y²+z²)² = x⁴+y⁴+z⁴+2(12)

(9)² = x⁴+y⁴+z⁴+24

81 = x⁴+y⁴+z⁴+24

81-24 = x⁴+y⁴+z⁴

57 = x⁴+y⁴+z⁴

A = 0, B = 2, C = 57

Subhotosh and Dr. Peterson, thank you very much for your help!

I think you typed that wrong, and it should be (x²+y²+z²)² = x⁴+y⁴+z⁴+B{(xy)²+(yz)²+(zx)²}.

Yes, I made a mistake there...

These may help as a start:

Thanks for the links. I'll check them out ?