If alpha and beta are the …

Here α and β are the zeros of polynomial f(x) = x² - px + q
So a=1,b=-p,c=q
Sum of the zeroes α + β={tex}-\frac ba{/tex} = p
Product of the zeroes  αβ=q
 {tex}\frac{{{\alpha ^2}}}{{{\beta ^2}}} + \frac{{{\beta ^2}}}{{{\alpha ^2}}}{/tex}
{tex}= \frac{{{\alpha ^4} + {\beta ^4}}}{{{\alpha ^2}{\beta ^2}}}{/tex}
{tex}=\frac{\left(\mathrm\alpha^2+\mathrm\beta^2\right)^2-2\left(\mathrm{αβ}\right)^2}{\left(\mathrm{αβ}\right)^2}=\frac{\{(\mathrm\alpha+\mathrm\beta)^2-2\mathrm{αβ}\}^2-2\left(\mathrm{αβ}\right)^2}{\left(\mathrm{αβ}\right)^2}{/tex}

{tex}=\frac{(\mathrm p^2-2\mathrm q)^2-2\mathrm q^2}{\mathrm q^2}=\frac{\mathrm p^4+4\mathrm q^2-4\mathrm p^2\mathrm q-2\mathrm q^2}{\mathrm q^2}=\frac{\mathrm p^4+2\mathrm q^2-4\mathrm p^2\mathrm q}{\mathrm q^2}{/tex}

{tex}=\frac{\mathrm p4}{\mathrm q^2}-\frac{4\mathrm p^2\mathrm q}{\mathrm q^2}+\frac{2\mathrm q^2}{\mathrm q^2}=\frac{\mathrm p4}{\mathrm q^2}-\frac{4\mathrm p^2}{\mathrm q}+2=\mathrm{RHS}{/tex}
Hence, proved.