Question

If x+y+ z = 8 and xy +yz+zx = 20 find the value of x3+y3+z3 -3xyz.

Answer 1

$Given:\ x+y+z=8 \\ xy+yz+zx=20 \\ To \: find\: x^{3}+y^{3}+z^{3}-3xyz =?$

Answer:

$x+y+z=8 --------eq 1 \\ xy+yz+zx= 20 ----eq2 \\ squaring\:both\: side \\ x^{2}+y^{2} +z^{2}+2xy+2yz +2zx \\ x^{2}+y^{2}+z^{2}+2(xy+yz +zx ) \\ x^{2}+y^{2}+z^{2}+2(20)=64 \\ x^{2}+y^{2}+z^{2}= 64-40 =24 \\by:\ eq2 \\ x^{3}+y^{3}+z^{3}-3xyz(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx) \\ x^{3}+y^{3}+ z^{3}-3xyz=8(24-20) \\ x^{3}+y^{3}+z^{3}+3xyx=8(4) \\ x^{3}+y^{3}+z^{3}-3xyz= 32 Answer.$

Answer 2

ANSWER:

• Value of => x³+y³+z³-3xyz is 32

GIVEN:

• x+y+z = 8
• xy+yz+zx = 20

TO FIND:

• x³+y³+z³-3xyz

SOLUTION:

Formula:

=> x³+y³+z³-3xyz = (x+y+z)[x²+y²+z²-(xy+yz+zx)]

Here:

x+y+z = 8

xy+yz+zx = 20

Now:

=> x+y+z = 8

Squaring both sides we get;

=>(x+y+z)² = (8)²

=> x²+y²+z²+2(xy+yz+zx) = 64

Putting (xy+yz+zx) = 20

=> x²+y²+z² +2(20) = 64

=> x²+y²+z² = 64- 40

=> x²+y²+z² = 24

=> x³+y³+z³-3xyz = (x+y+z)[x²+y²+z²-(xy+yz+zx)]

Putting values we get;

=> x³+y³+z³-3xyz = (8)[(24)-(20)]

=> x³+y³+z³-3xyz = 8(4)

=> x³+y³+z³-3xyz = 32