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Dear student,

The solution to your question no.1 has been provided below;
$Q1.$
$$\begin{gathered} \mathrm{R}.\mathrm{H}.\mathrm{S} = \frac{1}{2}\left(x+y+z\right)\left[{\left(x-y\right)}^2+{\left(y-z\right)}^2+{\left(z-x\right)}^2\right] \sin ce: [{(a-b)}^2=a^2+b^2-2ab] \\ = \frac{1}{2}\left(x+y+z\right)\left[x^2+y^2-2xy+y^2+z^2-2yz+z^2+x^2-2xz\right] \\ = \frac{1}{2}\left(x+y+z\right)\left[2x^2+2y^2+2z^2-2xy-2yz-2xz\right] \\ = \frac{1}{2}\times 2\left(x+y+z\right)\left[x^2+y^2+z^2-xy-yz-xz\right] \\ = \left(x+y+z\right)\left[x^2+y^2+z^2-xy-yz-xz\right] \\ = x\left[x^2+y^2+z^2-xy-yz-xz\right]+y\left[x^2+y^2+z^2-xy-yz-xz\right]+z\left[x^2+y^2+z^2-xy-yz-xz\right] \\ = x^3+\cancel{x{y}^2}+\cancel{x{z}^2}-\cancel{x^{2}y}-xyz-\cancel{x^{2}z}+\cancel{y{x}^2}+y^3+\cancel{y{z}^2}-\cancel{x{y}^2}-\cancel{y^{2}z}-xyz+\cancel{x^{2}z}+\cancel{y^{2}z}+z^3-xyz-\cancel{y{z}^2}-\cancel{x{z}^2} \\ = x^3+y^3+z^3-3xyz \\ \mathrm{Therefore}; \mathrm{L}.\mathrm{H}.\mathrm{S} = \mathrm{R}.\mathrm{H}.\mathrm{S} \end{gathered}$$

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