If 4x^2+y^2=17 and xy=2 , where x,y>0 , then find the value of 8x^3+y…

The correct answer is Option (4) → 65

Here \(8x^3+y^3=(2x)^3+y^3\), a sum of two cubes. With \(a=2x\) and \(b=y\), first find \(a+b\), then use \(a^3+b^3=(a+b)^3-3ab(a+b)\).

• \((2x+y)^2=4x^2+y^2+4xy=17+8=25\)

• \(x,y>0\), so \(2x+y=5\)

• \(2x\times y=2xy=4\)

• \(8x^3+y^3=5^3-3\times4\times5=125-60=65\)