If \(a(a+b+c)^2=1792\), \(b(a+b+c)^2=1536\) and \(c(a+b+c)^2=768\), then find the value of \(b^2+11\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 47
Adding the three equations takes \((a+b+c)^2\) common, so the left side becomes \((a+b+c)^3\). Its cube root gives \(a+b+c\), and then \(b\) follows from the second equation.
\((a+b+c)(a+b+c)^2=1792+1536+768\)
\((a+b+c)^3=4096\), so \(a+b+c=16\)
\(b\times16^2=1536\), so \(b=\displaystyle\frac{1536}{256}=6\)
\(b^2+11=36+11=47\)