Question

Find the value of (x-a)^3+(x-b)^3+(x-c)^3-3(x -a)(x -b)(x -c) when a+b+c=3x.

Answer 1

Question:

Find the value of (x-a)^3+(x-b)^3+(x-c)^3-3(x -a)(x-b)(x -c) when a+b+c=3x.

Answer:

We will use the following identity:

$\\x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

$(x-a)^3+(x-b)^3+(x-c)^3-3(x -a)(x-b)(x -c)=(x-a+x-b+x-c){(x-a)^2+(x-b)^2+(x-c)^2-(x-a)(x-b)(x-c)} \\\\\ \mapsto(3x-a-b-c){(x-a)^2+(x-b)^2+(x-c)^2-(x-a)(x-b)(x-c)} \\\\\ \mapsto~Replacing ~3x~ by (a+b+c) \\\\\ \mapsto(a+b+c-a-b-c){(x-a)^2+(x-b)^2+(x-c)^2-(x-a)(x-b)(x-c)}\\\\\ \mapsto \cancel{(a+b+c)} -\cancel{(a-b+c)}{(x-a)^2+(x-b)^2+(x-c)^2-(x-a)(x-b)(x-c)} \\\\\ \mapsto0 \times {(x-a)^2+(x-b)^2+(x-c)^2-(x-a)(x-b)(x-c)}= 0$

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