If x + y + z = 6, x² + y² + z² = 8, and x³ + y³ + z³ = 5, what is x⁴ + y⁴ + z⁴?

Let A=x+y+z, B=xy+yz+zx, C=xyz then x,y,z solve p³−Ap²+Bp−C=0

Multiply by pⁿ‾³ : pⁿ−Apⁿ‾¹+Bpⁿ‾²−Cpⁿ‾³=0

Set p=x,y,z and sum → s(n)=As(n−1)−Bs(n−2)+Cs(n−3) where s(n)=xⁿ+yⁿ+zⁿ

s(0)=3, s(1)=A, s(2)=A²−2B so s(n) can be found recursively as a function of A,B,C

In particular, s(3) = A(A²−2B)−BA+3C = A³−3AB+3C

Given s(1)=6, s(2)=5, s(3)=8, A,B,C come from solving sequentially the equations

s(1)=A, s(2)=A²−2B, s(3)=A³−3AB+3C → A=6, B=14, C=41/3

Recursion is now s(n)=6s(n−1)−14s(n−2)+(41/3)s(n−3)

s(4) = 6*5 − 14*8 + (41/3)*6 = 0

Some others are s(5)=118/3, s(6)=913/3, s(7)=3826/3

Given x + y + z = 6

(x + y + z) ² = 36

x ² + y ² + z ² + 2 (xy + xz + yz) = 36 --------------(1)

Now x ² + y ² + z ² = 8

implies xy + xz + yz = 14

or (xy + xz + yz) ² = (14) ²

(xy) ² + (xz) ² + (yz) ² + 2 xyz (x + y + z) = 196

(xy) ² + (xz) ² + (yz) ² + 12 xyz = 196 ---------- (2)

using identity x ³ + y ³ + z ³ - 3xyz = (x + y + z) (x ² + y ² + z ²-xy -xz- yz)

= 6 × [8 - (14)]

= (- 36)

Now x ³ + y ³ + z ³ = 5

or 5 - 3xyz = (- 36)

3xyz = 41

xyz = (41/3)

put value of xyz = (41/3) into (2) to get (xy) ² + (xz) ² + (yz) ² + (12)(41/3) = 196

(xy) ² + (xz) ² + (yz) ² = 32

Because x ² + y ² + z ² = 8

So (x ² + y ² + z ²) ² = 64

x ^ 4 + y ^ 4 + z ^ 4 +2 {(xy) ² + (xz) ² + (yz) ²} = 64

So x ^ 4 + y ^ 4 + z ^ 4 = 64 - 2 (32) = 0

If x = 8 and z = a million, what's the fee of y? x = 6(z - 5) - (x - y)? If x = 8 and z = a million, what's the fee of y? x = 6(z - 5) - (x - y) 8= 6(a million-5) -(8-y) 8 = -24 -8 +y y = 40 ..............Ans D A. y = 28 B. y = 32 C. y = -24 D. y = 40 ..................Ans ______________________________________... <A and <B are complementary. If <A is seventy 5°, what's the degree of <B? A. 15° B. seventy 5° C. 27° D. one hundred and five° answer A+B = ninety seventy 5 +B = ninety B= ninety-seventy 5 = 15 ......................Ans A

Edit: Steps are rewritten for clarity.

x^3 + y^3 = (x + y) [(3/2) (x^2 + y^2) - (1/2) (x + y)^2]

=> 5 - z^3 = (6 - z) [(3/2) (8 - z^2) - (1/2) (6 - z)^2]

=> 5 - z^3 = (6 - z) (- 2z^2 + 6z - 6)

=> 5 - z^3 = 2z^3 - 18z^2 + 42z - 36

=> 3z^3 = 18z^2 - 42z + 41

=> 3z^4 = 18z^3 - 42z^2 + 41z ... ( 1 )

Similarly,

3x^4 = 18x^3 - 42x^2 + 41x ... ( 2 ) and

3y^4 = 18y^3 - 42y^2 + 41y ... ( 3 )

Adding (1 ), ( 2 ) and ( 3 ),

3 (x^4 + y^4 + z^4)

= 18 (x^3 + y^3 + z^3) - 42 (x^2 + y^2 + z^2) + 41 (x + y + z)

= 18 * 5 - 42 * 8 + 41 * 6 = 0

=> x^4 + y^4 + z^4 = 0.

Since x + y + z = 6, squaring both sides yields

x² + y² + z² + 2(xy + xz + yz) = 6²

==> 8 + 2(xy + xz + yz) = 36

==> xy + xz + yz = 14.

Since x + y + z = 6, cubing both sides yields

x³ + y³ + z³ + 3(x²y + x²z + xy² + y²z + xz² + yz²) + 6xyz = 6³

==> 5 + 3(x²y + x²z + xyz + xy² + y²z + xyz + xz² + yz² + xyz) - 3xyz = 216

==> 3(x(xy + xz + yz) + y(xy + yz + xz) + z(xz + yz + xy)) - 3xyz = 211

==> 3(x + y + z)(xy + xz + yz) - 3xyz = 211

==> 3 * 6 * 14 - 3xyz = 211

==> xyz = 41/3

Squaring xy + xz + yz = 14:

x²y² + x²z² + y²z² + 2(x²yz + xy²z + xyz²) = 14²

==> x²y² + x²z² + y²z² + 2xyz(x + y + z) = 196

==> x²y² + x²z² + y²z² + 2 * 41/3 * 6 = 196

==> x²y² + x²z² + y²z² = 32.

Finally, squaring x² + y² + z² = 8:

x⁴ + y⁴ + z⁴ + 2(x²y² + x²z² + y²z²)= 64

==> x⁴ + y⁴ + z⁴ + 2 * 32 = 64

==> x⁴ + y⁴ + z⁴ = 0.

Sanity check (via wolfram alpha):

The given three equations has complex roots whose 4th powers do sum to 0...

I hope this helps!

1) By an algebraic identity, xy + yz + zx = (1/2)[(x+y+z)² - (x² + y² + z²)]

==> xy + yz + zx = (1/2)(36 - 8) = 14.

2) By another identity,

x³ + y³ + z³ - 3xyz = (x+y+z){(x² + y² + z²) - (xy+yz+zx)}

==> 5 - 3xyz = 6(8 - 14) = -36

==> xyz = 41/3

3) Also,

x²y² + y²z² + z²x² = (xy+yz+zx)² - 2xyz(x+y+z) = 196 - 2(41/3)(6) = 32

4) x⁴ + y⁴ + z⁴= (x² + y² + z²)² - 2(x²y² + y²z² + z²x²) = 64 - 64 = 0

2

11, 18 and 21 xoxoxoxoxoxo