Factor (1/16)x^6 + x^3 + 4 completely.

Step-by-step solution

1. 1

   Spot the hidden quadratic. The exponents are $6,3,0$ — each a multiple of $3$ — and $x^6=(x^3)^2$. So the expression is a quadratic in the single quantity $x^3$:

   $\frac{1}{16}\left(x^3\right)^2+1\cdot x^3+4$

2. 2

   Substitute u = x^3 and clear the fraction. With $u=x^3$ the expression is $\tfrac{1}{16}u^2+u+4$. Factoring out $\tfrac{1}{16}$ — equivalently multiplying inside by $16$ — gives whole-number coefficients:

   $\frac{1}{16}\left(u^2+16u+64\right)$

   Check the bookkeeping: $16\times1=16$ and $16\times4=64$ ✓.

3. 3

   Recognise the perfect square. For $u^2+16u+64$, note $64=8^2$ and $2\times8=16$ matches the middle coefficient, so

   $u^2+16u+64=(u+8)^2$

   (Equivalently the discriminant $16^2-4\cdot64=0$, confirming a repeated root.)

4. 4

   Substitute back.

   $\frac{1}{16}x^6+x^3+4=\frac{1}{16}\left(x^3+8\right)^2$

5. 5

   Push the factoring further with the sum of cubes. Since $x^3+8=x^3+2^3$, the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$ gives $x^3+8=(x+2)(x^2-2x+4)$, so squaring:

   $\frac{1}{16}x^6+x^3+4=\frac{1}{16}(x+2)^2\left(x^2-2x+4\right)^2$

   The quadratic $x^2-2x+4$ has discriminant $4-16=-12<0$, so it is irreducible over the reals and the factoring is complete.

6. 6

   Verify numerically. Comparing the original expression with both factored forms at $60$ random values of $x$ in $[-4,4]$ gives agreement to machine precision throughout ✓. As a spot check at $x=2$: original $=\tfrac{64}{16}+8+4=16$, and $\tfrac{1}{16}(8+8)^2=\tfrac{256}{16}=16$ ✓.