<s>[INST] Show that this expression is a perfect square? Show that this expression is a perfect square?
$(b^2 + 3a^2 )^2 - 4 ab*(2b^2 - ab - 6a^2)$
 [/INST] Observe that  $4ab(2b^2-6a^2)=8ab(b^2-3a^2)$
Using $(A+B)^2=(A-B)^2+4AB$
$$(b^2+3a^2)^2=(b^2-3a^2)^2+4(b^2)(3a^2)$$
$$\implies(b^2+3a^2)^2-8ab(b^2-3a^2)+4a^2b^2$$
$$=(b^2-3a^2)^2+12a^2b^2-8ab(b^2-3a^2)+4a^2b^2$$
$$=(\underbrace{b^2-3a^2}_p)^2 -2(\underbrace{b^2-3a^2}_p)(\underbrace{4ab}_q)+(\underbrace{4ab}_q)^2$$  
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