I would like to know the process of factoring this algebraic expression: 1) $64a^6+16a^3+1$
Thank you in advance!
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
set $$a^3=t$$ and factor $$64t^2+16t+1$$
answered Sep 11, 2017 at 11:22
-
$\begingroup$ how do you arrive at this answer: (2a+1)^2(4a^2-2a+1)^2 $\endgroup$ Commented Sep 11, 2017 at 11:28
-
$\begingroup$ $$64t^2+16t+1=64(t+\frac{1}{8})^2$$ $\endgroup$ Commented Sep 11, 2017 at 11:32
$\begingroup$
$\endgroup$
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.
answered Sep 11, 2017 at 11:34