Simplify (3x^2 + 4)^2 + (3x^2 - 4)^2 - 2(3x^2 + 4)(3x^2 - 4).

Step-by-step solution

1. 1

   Resist expanding — look at the shape first. Written out, this would be three quartic expansions and a lot of bookkeeping. But the three pieces are exactly $a^2$, $b^2$ and $-2ab$ for two repeated blocks, which is the signature of

   $a^2+b^2-2ab=(a-b)^2$

2. 2

   Substitute the blocks. Let

   $a=3x^2+4,\qquad b=3x^2-4$

   Then the expression is literally $a^2+b^2-2ab$, so it equals $(a-b)^2$. Treating each bracket as a single object is the move that saves all the work.

3. 3

   Compute $a-b$ and distribute the minus sign carefully.

   $a-b=(3x^2+4)-(3x^2-4)=3x^2+4-3x^2+4=8$

   The $3x^2$ terms cancel and the two constants add (because $-(-4)=+4$). Getting $0$ here instead of $8$ is the standard sign error.

4. 4

   Square the difference.

   $(a-b)^2=8^2=64$

   So the whole expression collapses to the constant $64$: it takes the same value for every real $x$, with no $x$ left in the answer.

5. 5

   Check at two different values of $x$. At $x=0$: $16+16-2(4)(-4)=16+16+32=64$. ✓ At $x=1$: $49+1-2(7)(-1)=49+1+14=64$. ✓ Two different inputs giving the same output is strong evidence the expression really is constant.