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06/07/2021 12:04 pm
Topic starter
Verify that:
x3 + y3 + z3 – 3xyz = (1/2) (x+y+z)[(x–y)2+(y–z)2+(z–x)2]
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06/07/2021 12:04 pm
We know that,
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]