Question

If x+y+z=0show that x² + y³+z³=3×yz

Answer 1

★ Given That :-

• x + y + z = 0

★ To Prove :-

• x³ + y³ + z³ = 3xyz

★ As We Know That :-

$\large{\bf\blue{\boxed{ {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) }}}$

Then,

On substituting the value of x + y + z = 0 in the above equation, we have

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

if we multiply any number with zero, it becomes zero.

=> x³ + y³ + z³ - 3xyz = 0

Therefore,

$\bf\pink{ {x}^{3} + {y}^{3} + {z}^{3} = 3xyz }$

L.H.S = R.H.S.

Hence, proved!!