Question

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

Answer 1

Given,

LHS : x³+y³+z³-3xyz

RHS : (x+y+z)[(x-y)²+(y-z)²+(z-x)²]

From RHS

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

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

=> (x+y+z)[2x²+2y²+2z²-2xy-2yz-2zx]

=> (x+y+z)(2)(x²+y²+z²-xy-yz-zx)

=> 2[x³+xy²+xz²-x²y-xyz-zx²+x²y+y³+yz²-xy²-y²z-xyz+x²z+y²z+z³-xyz-yz²-z²x]

=> 2[x³+y³+z³-xyz]

Hence proved!!

Answer 2

Step-by-step explanation:

x3+y3+z3−3xyz = (x+y+z)(x2+y2+z2–xy–yz–xz)

⇒ x3+y3+z3–3xyz = (1/2)(x+y+z)[2(x2+y2+z2–xy–yz–xz)]

= (1/2)(x+y+z)(2x2+2y2+2z2–2xy–2yz–2xz)

= (1/2)(x+y+z)[(x2+y2−2xy)+(y2+z2–2yz)+(x2+z2–2xz)]

= (1/2)(x+y+z)[(x–y)2+(y–z)2+(z–x)2]