Question

ɪ ᴡᴀɴᴛ ᴀɴꜱᴡᴇʀ ꜱᴛᴇᴘ ʙʏ ꜱᴛᴇᴘ ᴘʟꜱ.
ɴᴏ ɪʀʀᴇʟᴇᴠᴀɴᴛ ᴀɴꜱᴡᴇʀꜱ ᴘʟꜱ.​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

Verify that :

$\sf{x^{3}+y^{3}+z^{3}-3 x y z=\dfrac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]}$

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♣ ᴀɴꜱᴡᴇʀ :

In the given equation :

L.H.S =  $\sf{x^{3}+y^{3}+z^{3}-3 x y z}$

R.H.S = $\sf{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}$

Let us take $\sf{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}$ in L.H.S instead of  

R.H.S and $\sf{x^{3}+y^{3}+z^{3}-3 x y z}$  in R.H.S instead of L.H.S

Interchanging positions doesn't effect answer

Now, In the given equation (after interchanging) :

L.H.S =  $\sf{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}$

R.H.S = $\sf{x^{3}+y^{3}+z^{3}-3 x y z}$

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Solve $\sf{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}$ now :

$\sf{\dfrac{1}{2}\left(x+y+z\right)\left(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right)}$

$\sf{=\dfrac{1\cdot \left(x+y+z\right)\left(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right)}{2}}$

$\sf{=\dfrac{\left(x+y+z\right)\left(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right)}{2}}$

$\sf{=\dfrac{\left(x+y+z\right)\left(2x^2-2xy-2xz+2y^2+2z^2-2yz\right)}{2}}$

$\sf{=\dfrac{\left(x+y+z\right)\cdot \:2\left(x^2-xy-xz+y^2+z^2-yz\right)}{2}}$

$\sf{=\left(x+y+z\right)\left(x^2-xy-xz+y^2+z^2-yz\right)}$

$\sf{=\left(x+y+z\right)(x^2+ y^2 + z^2 - xy - yz- xz)}$

We know :

$\bf{x^3+ y^3 + z^3- 3xyz = (x + y + z)(x^2+ y^2 + z^2 - xy - yz- xz)}$

$\large\boxed{\sf{\left(x+y+z\right)(x^2+ y^2 + z^2 - xy - yz- xz)=\sf{x^{3}+y^{3}+z^{3}-3 x y z}}}$

L.H.S = R.H.S

Hence Proved !!!