Question 1205171
If {{{alpha}}}, {{{beta}}}, {{{gamma}}} are the roots of {{{x^3+px^2 + qx +r = 0}}}, then we know that 

{{{alpha + beta + gamma = -p}}},
{{{alpha*beta + beta*gamma + alpha*gamma = q}}}, and
{{{alpha*beta*gamma = -r}}}.

The 2nd equation above gives
{{{(alpha*beta + beta*gamma + alpha*gamma)^2 = q^2}}}.

But the left side of the preceding equation expands to 

{{{alpha^2*beta^2 + beta^2*gamma^2 + alpha^2*gamma^2 + 2(alpha*beta^2*gamma + alpha^2*beta*gamma + alpha*beta*gamma^2) = alpha^2*beta^2 + beta^2*gamma^2 + alpha^2*gamma^2 + 2*alpha*beta*gamma*(beta + alpha + gamma)}}}.

Hence, {{{alpha^2*beta^2 + beta^2*gamma^2 + alpha^2*gamma^2 + 2*alpha*beta*gamma*(beta + alpha + gamma) = alpha^2*beta^2 + beta^2*gamma^2 + alpha^2*gamma^2 + 2rp = q^2}}}.

Therefore, {{{alpha^2*beta^2 + beta^2*gamma^2 + alpha^2*gamma^2 = highlight(q^2 - 2pr)}}}