Show that this expression is a perfect square? Show that this expression is a perfect square?
$(b^2 + 3a^2 )^2 - 4 ab*(2b^2 - ab - 6a^2)$
 A: Here's a sketch of a how-to:


*

*Expand out: $9a^4+24a^3b+10a^2b^2-8ab^3+b^4$

*Dehomogenize: $9x^4+24x^3+10x^2-8x+1$

*Equate with $(Ax^2+Bx+C)^2$, solve for $A,B,C$

A: Observe that  $4ab(2b^2-6a^2)=8ab(b^2-3a^2)$
Using $(A+B)^2=(A-B)^2+4AB$
$$(b^2+3a^2)^2=(b^2-3a^2)^2+4(b^2)(3a^2)$$
$$\implies(b^2+3a^2)^2-8ab(b^2-3a^2)+4a^2b^2$$
$$=(b^2-3a^2)^2+12a^2b^2-8ab(b^2-3a^2)+4a^2b^2$$
$$=(\underbrace{b^2-3a^2}_p)^2 -2(\underbrace{b^2-3a^2}_p)(\underbrace{4ab}_q)+(\underbrace{4ab}_q)^2$$  
A: The expression equals to: $b^4\cdot \left(3t^2 + 4t - 1\right)^2$, with $t = \dfrac{a}{b}$. Now substitute $t$ back into the expression.
A: In fact, note that $(x + y - z)^2 = x^2 + y^2 + z^2 + 2xy - 2yz - 2xz$. Thus,
$$
(3a^2 + b^2)^2 - 8ab^3 + 4a^2b^2 + 24a^3b = 9a^4 + 10a^2b^2 + b^4 - 8ab^3 + 24a^3b
$$
$$
= 9a^4 + (16a^2b^2 - 6a^2b^2) + b^4 + 24a^3b - 8ab^3 
$$
$$
= 9a^4 + 16a^2b^2 + b^4 + 24a^3b - 6a^2b^2 - 8ab^3
$$
$$
= (3a^2)^2 + (4ab)^2 + (-b^2)^2 + 2(3a^2)(4ab) + 2(3a^2)(-b^2) + 2(4ab)(-b^2) 
$$
$$
= (3a^2 + 4ab - b^2)^2
$$
A: Once you factored it out and got $9a^4+24a^3b+10a^2b^2−8ab^3+b^4$, you'd suspect that this is the square of a sum. What are the parts of the sum? 
If we let b = 0, the sum is $9a^4$ so there should be a term $±3a^2$. If we let a = 0, the sum is $b^4$, so there should be a term $±b^2$. 
We know that $(x+y)^2=x^2 + 2xy + y^2$. Guessing that $x=3a^2$ and $24a^3b=2xy=6a^2y$, that would make y = 4ab. From the other side, if $x=±b^2$ and $2xy=-8ab^3=-/+2yb^2$, that makes $x = -b^2$ and $y = 4ab$. 
So an educated guess would be $(3a^2+4ab-b^2)^2$, which lucky enough produces the requested result. 
