Question

If:

x + y + z = 7

x² + y² + z² = 33

x³ + y³ + z³ = 145

What is:
x⁴ + y⁴ + z⁴​

Answer 1

Take some trinomial expansions to find some useful expressions and their values.

• 2nd degree expansion

`x + y + z = 7 \\\\ \implies (x+y+z)^2 = x^2+y^2+z^2+2(xy+xz+yz) = 7^2 \\\\ \implies xy+xz+yz = \frac{49-33}2 = 8`

• 3rd degree expansion

`x+y+z=7 \implies (x+y+z)^3 = 7^3 \\\\ \implies \\ 243 = x^3+y^3+z^3+3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) + 6xyz \\\\ \implies \\ x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 + 2xyz = \frac{343 - 145}3 = 66`

• 4th degree expansion

`x+y+z=7 \implies (x+y+z)^4 = 7^4 \\\\ \implies \\ 2401 = x^4+y^4+z^4+4(x^3y+x^3z+xy^3+y^3z+xz^3+yz^3) \\\\ ~~~~~~~~~~~~+ 6(x^2y^2+x^2z^2+y^2z^2) + 12(x^2yz+xy^2z+xyz^2) \\\\ \implies \\\\ x^4+y^4+z^4 = 2401 - 4(x^3y+x^3z+xy^3+y^3z+xz^3+yz^3) \\\\ ~~~~~~~~~~~~~~~~~~~~~~- 6(x^2y^2+x^2z^2+y^2z^2) - 12(x^2yz+xy^2z+xyz^2)`

Simplify:

`x^3y+x^3z+xy^3+y^3z+xz^3+yz^3 \\\\ ~~~~ = x^2 (xy+xz) + y^2 (xy+yz) + z^2 (xz + yz) \\\\ ~~~~ = x^2 (xy+xz+yz) + y^2 (xy+xz+yz) + z^2 (xy + xz + yz) - (x^2yz + xy^2z + xyz^2) \\\\ ~~~~ = (x^2+y^2+z^2) (xy+xz+yz) - xyz (x+y+z) \\\\ ~~~~ = 264 - 7xyz`

`x^2y^2+x^2z^2+y^2z^2 \\\\ ~~~~ = xy\cdot xy + xz \cdot xz + yz \cdot yz \\\\ ~~~~ = xy(xy+xz+yz) + xz(xy+xz+yz) + yz(xy+xz+yz) - 2(x^2yz+xy^2z+xyz^2) \\\\ ~~~~ = (xy+xz+yz)^2 - 2xyz(x+y+z) \\\\ ~~~~ = 64 - 14xyz`

`x^2yz+xy^2z+xyz^2 = xyz (x + y + z) = 7xyz`

Then the equation we got from the 4th degree expansion reduces to

`x^4+y^4+z^4 = 2401 - 4(264-7xyz) - 6(64-14xyz) - 12(7xyz) \\\\ ~~~~~~~~~~~~~~~~~= 961 + 28xyz`

and all we need now is the value of
`xyz`.

In the 3rd degree expansion, we have

`x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 \\\\ ~~~~ = x(xy + xz) + y(xy + yz) + z(xz + yz) \\\\ ~~~~ = x(xy+xz+yz) + y(xy+xz+yz) + z(xy+xz+yz) - 3xyz \\\\ ~~~~ = (x+y+z)(xy+xz+yz) - 3xyz \\\\ ~~~~ = 56 - 3xyz \\\\ \implies (56 - 3xyz) + 2xyz = 66 \\\\ \implies xyz = -10`

So, we end up with

`x^4+y^4+z^4 = 961 - 280 = \boxed{681}`