0
$\begingroup$

I would like to know the process of factoring this algebraic expression: 1) $64a^6+16a^3+1$

Thank you in advance!

$\endgroup$
1
$\begingroup$

set $$a^3=t$$ and factor $$64t^2+16t+1$$

$\endgroup$
  • $\begingroup$ how do you arrive at this answer: (2a+1)^2(4a^2-2a+1)^2 $\endgroup$ – user370026 Sep 11 '17 at 11:28
  • $\begingroup$ $$64t^2+16t+1=64(t+\frac{1}{8})^2$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 11 '17 at 11:32
0
$\begingroup$

Let $u = a^3$. Then $$64a^6 + 16a^3 + 1 = 64u^2 + 16u + 1 = (8u + 1)^2 = (8a^3 + 1)^2$$ Now use the sum of cubes formula $$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$ to factor the term inside the parentheses.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.