Question

please give the right answer to my question

please don't spam....its urgent ​

Answer 1

Answer:

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and bro I'm studying in 10th class CBSE

Answer 2

Step-by-step explanation:

To prove:

x^3 + y^3 + z^3 - 3xyz = (1/2)×(x+y+z)[(x-y)^2 + (y-z)^2 + (z-x)^2];

Proof: We will consider only RHS for the sake of our understanding.

Thus, LHS = (1/2)×(x+y+z)[(x-y)^2 + (y-z)^2 + (z-x)^2];

Using the identity, (a-b)^2 = a^2 + b^2 -2ab;

=(1/2)×(x+y+z)[(x^2+y^2-2xy)+ (z^2+y^2-2zy) + (z^2+x^2-2zx)];

=(1/2)×(x+y+z)[x^2+y^2-2xy+ z^2+y^2-2zy + z^2+x^2-2zx];

=(1/2)×(x+y+z)[2x^2 +2z^2+2y^2-2zy -2zx -2xy];

=(1/2)×(x+y+z) × 2[x^2+z^2+y^2-zy -zx -xy];

=(x+y+z)[x^2 +z^2+y^2-zy -zx -xy];

=[x^3 +xz^2+xy^2-xzy -zx^2 -x^2y] + [yx^2 +yz^2+y^3-zy^2 -zxy -xy^2] + [zx^2 +z^3+zy^2-z^2y -z^2x -zxy];

=x^3 -xzy+y^3-zxy+z^3 -zxy;

=x^3 +y^3 +z^3 -3zxy;

x^3 +y^3 +z^3 -3zxy = x^3 +y^3 +z^3 -3zxy;

Thus, LHS = RHS.

That's all.