Step 1

Consider the following problem \(\displaystyle{x}^{{3}}−{3}{x}^{{2}}+{27}\equiv{0}{\left(\text{mod}{225}\right)}\)

Prime factorization of 225 is \(\displaystyle{3}^{{2}}\times{5}^{{2}}\)

Now, \(\displaystyle{225}=\frac{{{3}^{{2}}\times{5}^{{2}}}}{{{x}^{{3}}-{3}{x}^{{2}}+{27}}}\) implies that the polynomial is reducible over \(\displaystyle{F}_{{3}}{\quad\text{and}\quad}{F}_{{5}}\)

So, there is a solution under 3 with x=0 and a solution under modulo 5 with x=1

Since the modulus is composite , so we solve the equation with \(\displaystyle\text{mod}{\left({3}^{{2}}\times{5}^{{2}}\right)}\)

Therefore, we have

x=0(mod 3)

x=1(mod 5)

Step 2

Now, using Chinese remainder theorem for

x=0(mod 3)

x=1(mod 5)

we get x=6

Consider the following problem \(\displaystyle{x}^{{3}}−{3}{x}^{{2}}+{27}\equiv{0}{\left(\text{mod}{225}\right)}\)

Prime factorization of 225 is \(\displaystyle{3}^{{2}}\times{5}^{{2}}\)

Now, \(\displaystyle{225}=\frac{{{3}^{{2}}\times{5}^{{2}}}}{{{x}^{{3}}-{3}{x}^{{2}}+{27}}}\) implies that the polynomial is reducible over \(\displaystyle{F}_{{3}}{\quad\text{and}\quad}{F}_{{5}}\)

So, there is a solution under 3 with x=0 and a solution under modulo 5 with x=1

Since the modulus is composite , so we solve the equation with \(\displaystyle\text{mod}{\left({3}^{{2}}\times{5}^{{2}}\right)}\)

Therefore, we have

x=0(mod 3)

x=1(mod 5)

Step 2

Now, using Chinese remainder theorem for

x=0(mod 3)

x=1(mod 5)

we get x=6