Here's an alternative method.
\(\displaystyle abc = 15, \text{ so, }abcd=15d,\text{ so, }\frac{abcd}{15}=d \quad \dots\quad(1).\)
Similarly,
\(\displaystyle \frac{abcd}{30}=a,\quad \dots\quad (2),\\ \displaystyle \frac{abcd}{10}=b,\quad \dots \quad(3),\\ \displaystyle \frac{abcd}{6}=c,\quad \dots \quad (4).\)
Adding (1), (2), (3) and (4),
\(\displaystyle a+b+c+d=abcd\left(\frac{1}{15}+\frac{1}{30}+\frac{1}{10}+\frac{1}{6}\right)=abcd\left(\frac{11}{30}\right)\quad \dots \quad (5). \)
Multiplying the original four equations,
\(\displaystyle a^{3}b^{3}c^{3}d^{3}=15\times 30\times10\times6=2^{3}\times3^{3}\times5^{3},\\ \text{ so }abcd=2\times 3 \times 5 = 30.\)
Then, from (5),
\(\displaystyle a+b+c+d=30.\left(\frac{11}{30}\right)=11.\)
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