Factor the expression 75a⁶ − 147b¹⁶ completely.

Step-by-step solution

1. 1

   Always take out the greatest common factor first. As written, neither $75a^6$ nor $147b^{16}$ is a perfect square, so the difference-of-squares pattern does not apply yet. Factor the coefficients:

   $75=3\cdot 25,\qquad 147=3\cdot 49$

   The variables share nothing — $a^6$ and $b^{16}$ have no letter in common — so the GCF is just the number $3$:

   $75a^6-147b^{16}=3\left(25a^6-49b^{16}\right)$

2. 2

   Check whether each term inside the bracket is a perfect square. This is what the GCF step was for. A monomial is a perfect square when its coefficient is a square and its exponents are even:

   $25a^6=(5a^3)^2\qquad\text{since } 5^2=25 \text{ and } (a^3)^2=a^6$

   $49b^{16}=(7b^8)^2\qquad\text{since } 7^2=49 \text{ and } (b^8)^2=b^{16}$

   Halving the exponent is the whole trick: $6/2=3$ and $16/2=8$.

3. 3

   Apply the difference-of-squares identity. With $u=5a^3$ and $v=7b^8$,

   $u^2-v^2=(u-v)(u+v)$

   gives

   $25a^6-49b^{16}=(5a^3-7b^8)(5a^3+7b^8)$

4. 4

   Reassemble with the GCF you set aside. Do not lose the $3$ — it is the single most common slip in this type:

   $75a^6-147b^{16}=3(5a^3-7b^8)(5a^3+7b^8)$

5. 5

   Confirm nothing factors further, then spot-check. A sum of squares like $5a^3+7b^8$ never factors over the reals, and $5a^3-7b^8$ is not a difference of squares because $5$ and $7$ are not perfect squares and the exponent $3$ is odd. To be safe, expand the answer back:

   $3\left[(5a^3)^2-(7b^8)^2\right]=3\left(25a^6-49b^{16}\right)=75a^6-147b^{16}$

   which matches the original expression.