Step-by-step explanation:
[tex] {c}^{2} - 6c + 9 \\ = {c}^{2} - 2 \times 3 \times c + {3}^{2} \\ = (c - 3)^{2} \\ [/tex]
Therefore, [tex] c^2 - 6c +9 [/tex] is a perfect square.
Answer:(c-3)(c-3) yes
C²-6c-9
(c²-3c-)(3c+9)
c(c-3)-3(c-3)
(c-3)(c-3)
