Question

if x+y+z = 0 , show that x^3+ y^3+ z^3 = 3xyz​

Answer 1

Answer:

Step-by-step explanation:

x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx). = (x + y + z) (x^2 + y^2 + z^2 – xy – yz – zx).

When x+y+z=0

x^3 + y^3 + z^3 – 3xyz = (0) (x^2 + y^2 + z^2 – xy – yz – zx)

x^3 + y^3 + z^3 -3xyz = 0

x^3 + y^3 + z^3 = 3xyz(hence proved)

MARK AS BEAINLIST

Answer 2

$\bf{\blue{\bigstar}}$ Formula Used :

$\boxed{\bold{\bf{\red{(x+y)^3=x^3+y^3+3xy(x+y)}}}}$

$\bf{\blue{\bigstar}}$ Proof :

$\bold{x+y+z=0\ [ \;Given\ ]}\\\\\implies \bold{x+y=-z\;...(1)}\\\\\bold{Now\ cubing\ on\ both\ sides\ , we\ get\ ,}\\\\\implies \bold{(x+y)^3=(-z)^3}\\\\\implies \bold{x^3+y^3+3xy(x+y)=-z^3}\\\\\implies \bold{x^3+y^3+3xy(-z)=-z^3\; [\; From\ (1)]}\\\\\implies \bold{x^3+y^3+z^3-3xyz=0}\\\\\implies \boxed{\bold{\bf{\red{x^3+y^3+z^3=3xyz}}}}$

Hence showed