Question:
Verify that x³+y³+z³-3xyz=1/2(x+y+z)[(x-y)²+(y-z)²+(z-x)²]
Mam can you tell how to solve this question ❓🙏🙏
Answer:
RHS:
(1/2) * (x + y + z) * [(x - y)2 + (y - z)2 + (z - x)2]
= (1/2) * (x + y + z) * [x2 + y2 - 2xy + y2 + z2 - 2yz + z2 + x2 - 2zx]
= (1/2) * (x + y + z) * [2x2 + 2y2 + 2z2 – 2xy - 2yz - 2zx]
= (1/2) * 2 * (x + y + z) * [x2 + y2 + z2 – xy - yz - zx]
= (x + y + z) * [x2 + y2 + z2 – xy - yz - zx]
= x3 + y3 + z3 – 3xyz
[Using a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc - ca)]
= LHS
So, x3 + y3 + z3 – 3xyz = (1/2) * (x + y + z) * [(x - y)2 + (y - z)2 + (z - x)2] = x3 + y3 + z3 – 3xyz
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