Question

Hallar la suma de: x^3+y^3+z^3=?
{(x+y+z=5@x^2+y^2+z^2=3^2@1/x+1/y+1/z=1/2)

Answer 1

1) $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2}\iff 2(xy+xz+yz)=xyz$

2) 
      $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\\ \\ 25=9+xyz\\ \\ xyz=16$

3) $xy+xz+yz=8$

4) 
              $(x+y+z)^3=x^3+y^3+z^3+3(x+y)(x+z)(y+z)\\ \\ 125=x^3+y^3+z^3+3(5-z)(5-y)(5-x)\\ \\ x^3+y^3+z^3=125+(x-5)(y-5)(z-5) \\ \\ x^3+y^3+z^3=125+(xy-5x-5y+25)(z-5)\\ \\ x^3+y^3+z^3=125+xyz-5xz-5yz+25z-5xy+25x+25y-125\\ \\ x^3+y^3+z^3=xyz-5(xz+yz+xy)+25(x+y+z)\\ \\ x^3+y^3+z^3=16-5(8)+25(5)\\ \\ \\ \boxed{x^3+y^3+z^3=101}$

Answer 2

Respuesta:

Explicación paso a paso:

1st eq. => x+y+z = 5

=> (x+y+z)² = 5²

=> x² + y² + z² + 2(xy+xz+yz) = 25

2nd eq. => x² + y² + z² = 3² = 9

3rd eq. · xyz => yz + xz + xy = \frac{xyz}{2}

1st eq. => x² + y² + z² + 2(xy+xz+yz) = 25

=> 9 + 2(\frac{xyz}{2} ) = 25

=> 2(\frac{xyz}{2} ) = 16

=> xyz = 16

3rd eq. => yz + xz + xy = \frac{xyz}{2}

=> yz + xz + xy = \frac{16}{2} = 8

Note:

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

= x³ + y³ + z³ + 3xy(x + y) + 3yz(y + z) + 3xz(x + z) + 6xyz

Use:

xyz = 16 => xy = \frac{16}{z} => xz = \frac{16}{y} => yz = \frac{16}{x}

x + y + z = 5 => x + y = 5 - z => x + z = 5 - y => y + z = 5 - x

Substitute:

(x + y + z)³ = x³ + y³ + z³ + 3xy(x + y) + 3yz(y + z) + 3xz(x + z) + 6xyz =

5³ = x³ + y³ + z³ + 3 · \frac{16}{z} · (5 - z) + 3 · \frac{16}{x} · (5 - x) + 3 · \frac{16}{y} · (5 - y) + 6 · 16

=> 125 = x³ + y³ + z³ + \frac{48}{z} · (5 - z) + \frac{48}{x} · (5 - x) + \frac{48}{y} · (5 - y) + 96

=> 125 = x³ + y³ + z³ + \frac{48 · 5}{z} - 48 + \frac{48 · 5}{x} - 48 + \frac{48 · 5}{y} - 48 + 96

=> 125 = x³ + y³ + z³ + \frac{240}{z} + \frac{240}{x} + \frac{240}{y} - 48 - 48 - 48 + 96

=> 125 = x³ + y³ + z³ + \frac{240 · xy}{z · xy} + \frac{240 · yz}{x · yz} + \frac{240 · xz}{y · xz} - 48

=> 125 = x³ + y³ + z³ + \frac{240xy}{xyz} + \frac{240yz}{xyz} + \frac{240xz}{xyz} - 48

=> 125 = x³ + y³ + z³ + \frac{240xy + 240yz + 240xz}{xyz} - 48

=> 125 = x³ + y³ + z³ + 240 · \frac{xy + yz + xz}{xyz} - 48

Use:

xy + yz + xz = 8

xyz = 16

Substitute:

=> 173 = x³ + y³ + z³ + 240 · \frac{xy + yz + xz}{xyz}

=> 173 = x³ + y³ + z³ + 240 · \frac{8}{16}

=> 173 = x³ + y³ + z³ + 240 · \frac{1}{2}

=> 173 = x³ + y³ + z³ + 120

=> 53 = x³ + y³ + z³