Question

Q12 and Q13
Step by step solution plz​

Answer 1

Answer:

Step-by-step explanation:

x³+y³+z³-3xyz=1/2(x+y+z)[(x-y)²+(y-z)²+(z-x)²]

Taken RHS

   =    1/2(x+y+z)[x²+y²-2xy+y²+z²-2yz+z²+x²-2zx]

   =   (x+y+z)[x²+y²+z²-xy-yz-zx]

ON MULTIPLYING WE GET

    =   X³+Y³+Z³-3XYZ

Ques 13 Given That

If x+y+z=0 then show that x³+y³+z³=3xyz

From above equetion  We Know that

x³+y³+z³-3xyz=1/2(x+y+z)[(x-y)²+(y-z)²+(z-x)²]

Put x+y+z=0

so RHS become 0 and

x³+y³+z³-3xyz=0

x³+y³+z³=3xyz