If \(9x^2+y^2=37\) and \(xy=2\), where \(x,y>0\), then find the value of \(27x^3+y^3+50\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 267
Here \(27x^3+y^3=(3x)^3+y^3\), a sum of two cubes. With \(a=3x\) and \(b=y\), first find \(a+b\), then use \(a^3+b^3=(a+b)^3-3ab(a+b)\).
\((3x+y)^2=9x^2+y^2+6xy=37+12=49\)
\(x,y>0\), so \(3x+y=7\)
\(3x\times y=3xy=6\)
\(27x^3+y^3=7^3-3\times6\times7=343-126=217\)
\(27x^3+y^3+50=217+50=267\)