Question

Hii please solve this ​

Answer 1

Answer:

SOlving R.H.S

Step-by-step explanation:

Using $(a-b)^{2}$ = $a^{2}$+$b^{2}$-2ab

= $\frac{1}{2}(x+y+z)[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}]$

=$\frac{1}{2}(x+y+z)[(x^{2}+y^{2} -2xy )+(y^{2} +z^{2}-2yz )+(z^{2}+x^{2}-2xz)]$

= $\frac{1}{2}(x+y+z)[2x^{2}+2y^{2}+2z^{2} -2xy-2yz -2xz]$

= $\frac{1}{2}(x+y+z)2[x^{2}+y^{2}+z^{2} -xy-yz -xz]$

= $(x+y+z)[x^{2}+y^{2}+z^{2} -xy-yz -xz]$

We know $x^{3} +y^{3} + z^{3}-3xyz = (x+y+z)[x^{2}+y^{2}+z^{2} -xy-yz -xz]$

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

= L.H.S (prooved)