Suppose that RSA is used to send a message \(m\) to three recipients, who have
relatively prime encryption moduli \(n_{1}, n_{2}\), and \(n_{3} .\) All three
recipients use the same encryption exponent \(e=3\), a once-popular choice as it
makes encryption very fast. Show that someone who intercepts all three
encrypted messages \(c_{1}=m^{3}\) \(\bmod n_{1}, c_{2}=m^{3} \bmod n_{2}\), and
\(c_{3}=m^{3} \bmod n_{1}\) can efficiently decipher \(m .\) Hint: The Chinese
remainder theorem implies that you can efficiently find a \(c\) such that
\(c=c_{1} \bmod n_{1}, c=c_{2} \bmod n_{2}\), and \(c=c_{3} \bmod n_{3} .\) Assume
this, and show that it implies \(c=m^{3} \bmod n_{1} n_{2} n_{3} .\) Then note
\(m^{3}