Question

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

Answer 1

As we know the formula of

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

If x + y + z = 0 then

x³ + y³ + z³ - 3xyz = 0 ( x²+y² + z² -xy -yz -zx)

x³ + y³ + z³ -3xyz = 0

Tranpose -3xyz to RHS

x³ + y³ + z³ = 3xyz

Hence proved

By the formula of x³ + y³ + z³ - 3xyz = (x + y + z ) ( x² + y² + z² -xy - yz - zx ) we can prove

Know more :-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc

a³ + b³ = ( a+b) (a²-ab+b²)

a³-b³ = (a - b) (a² + ab + b²)

Answer 2

$\huge\mathfrak\red{Solution}$

If x + y + z = 0.

show that :-

${x}^{3} + {y}^{3} + {z}^{3} = 3xyz$

We know that :-

${x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z) \: ( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)$

Now put 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

Hope this helps you.

# By Sparkly Princess