[tex]\begin{aligned}&\bullet\ a=\bf\dfrac{1}{5}\\&\bullet\ b=\bf243\\&\bullet\ c=\bf\dfrac{54}{5}\end{aligned}[/tex]
Explanation
The first 3 terms in the expansion of [tex](3 - ax)^5[/tex], in ascending powers of [tex]x[/tex], can be written in the form of [tex]b - 81x + cx^2[/tex].
So, expanding the first 3 terms of [tex](3 - ax)^5[/tex] using the binomial expansion, we get:
[tex]\begin{aligned}&\binom{5}{0}\cdot3^5(-ax)^0\\&+\binom{5}{1}\cdot3^4(-ax)^1\\&+\binom{5}{2}\cdot3^3(-ax)^2\\\\{=\ }&\frac{5!}{0!(5-0)!}\cdot243(1)\\&+\frac{5!}{1!(5-1)!}\cdot81(-ax)\\&+\frac{5!}{2!(5-2)!}\cdot27(a^2x^2)\\\\{=\ }&\frac{\cancel{5!}}{\cancel{5!}}\cdot243-\frac{5!}{4!}\cdot81ax+\frac{5!}{2\cdot3!}\cdot27a^2x^2\\\\{=\ }&243-\frac{5\cdot\cancel{4!}}{\cancel{4!}}\cdot81ax+\frac{5\cdot4\cdot\cancel{3!}}{2\cdot\cancel{3!}}\cdot27a^2x^2\\\\{=\ }&\boxed{\ 243-405ax+270a^2x^2\ }\end{aligned}[/tex]
Hence,
[tex]\begin{aligned}&b - 81x + cx^2=243-405ax+270a^2x^2\\&\Rightarrow\begin{cases}-405a=-81\\\Rightarrow a=\dfrac{-81}{-405}=\bf\dfrac{1}{5}\\\\b=\bf243\\\\c=270a^2=270\left(\frac{1}{5}\right)^2\\\Rightarrow c=\dfrac{270}{25}=\bf\dfrac{54}{5}\end{cases}\end{aligned}[/tex]
CONCLUSION
[tex]\begin{aligned}\therefore\quad&\bullet\ a=\bf\dfrac{1}{5}\\&\bullet\ b=\bf243\\&\bullet\ c=\bf\dfrac{54}{5}\end{aligned}[/tex]
[answer.2.content]
